问题

You are supposed to provide a solution to a given problem.

Problem:
Is there an eigenbasis for the identity matrix I_n?
Solution:

答案

模型	模型答案	正确答案
金声玉亮2.0	Yes, there is an eigenbasis for the identity matrix $I_n$. Explanation: An eigenbasis of a matrix is a basis for the vector space consisting of eigenvectors of that matrix. For a matrix to have an eigenbasis, it needs to be diagonalizable, meaning it can be expressed as $A = PDP^{-1}$, where $D$ is a diagonal matrix whose diagonal entries are the eigenvalues of $A$, and $P$ is the matrix whose columns are the eigenvectors of $A$. The identity matrix $I_n$ is a diagonal matrix with all diagonal entries equal to 1. Therefore, its eigenvalues are all 1, and any non-zero vector is an eigenvector of $I_n$ since $I_n v = v$ for any non-zero vector $v$. As a result, any basis of the vector space can be chosen as an eigenbasis for $I_n$, since all vectors in the basis will be eigenvectors of $I_n$. In summary, any basis of the vector space $\mathbb{R}^n$ (or whatever the underlying vector space is) can serve as an eigenbasis for the identity matrix $I_n$.	True
deepseekr1	The identity matrix $ I_n $ does have an eigenbasis. Reasoning: 1. Eigenvectors of $ I_n $: Every non-zero vector $ v $ in the vector space is an eigenvector of $ I_n $, since $ I_n v = v $, with eigenvalue $ 1 $. 2. Eigenspace: The eigenspace for the eigenvalue $ 1 $ is the entire vector space, as $ (I_n - I_n)v = 0 $ holds for all $ v $. Thus, the geometric multiplicity of the eigenvalue $ 1 $ is $ n $. 3. Diagonalizability: A matrix is diagonalizable if its eigenvectors form a basis (an eigenbasis). Since every basis of the vector space consists of eigenvectors of $ I_n $, such a basis exists. For example, the standard basis $ \{e_1, e_2, \dots, e_n\} $ is already an eigenbasis. Conclusion: Yes, there exists an eigenbasis for $ I_n $. In fact, every basis of the vector space is an eigenbasis for the identity matrix. $\boxed{\text{Yes}}$	True
gpto1	Yes, there is an eigenbasis for the identity matrix $ I_n $. In fact, every non-zero vector in $ \mathbb{R}^n $ (or $ \mathbb{C}^n $) is an eigenvector of the identity matrix, corresponding to the eigenvalue $ \lambda = 1 $. ### Explanation: The identity matrix $ I_n $ is a square $ n \times n $ matrix with ones on the main diagonal and zeros elsewhere: \[ I_n = \begin{pmatrix} 1 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \\ \end{pmatrix} \] An eigenvector $ \mathbf{v} $ of a matrix $ A $ is a non-zero vector that satisfies the equation: \[ A\mathbf{v} = \lambda \mathbf{v} \] where $ \lambda $ is the eigenvalue corresponding to $ \mathbf{v} $. #### For the Identity Matrix: 1. Eigenvalue Determination: Let's find the eigenvalues by solving the characteristic equation: \[ \det(I_n - \lambda I_n) = 0 \] Simplify $ I_n - \lambda I_n = (1 - \lambda)I_n $. The determinant of $ (1 - \lambda)I_n $ is: \[ \det((1 - \lambda)I_n) = (1 - \lambda)^n \det(I_n) = (1 - \lambda)^n (1) = (1 - \lambda)^n \] Setting the determinant to zero: \[ (1 - \lambda)^n = 0 \implies \lambda = 1 \] So, the only eigenvalue of $ I_n $ is $ \lambda = 1 $, and its algebraic multiplicity is $ n $. 2. Eigenvector Determination: For $ \lambda = 1 $, the eigenvector equation becomes: \[ I_n \mathbf{v} = \mathbf{v} \] Which simplifies to: \[ \mathbf{v} = \mathbf{v} \] This equation is always true for any vector $ \mathbf{v} $ in $ \mathbb{R}^n $ (or $ \mathbb{C}^n $). However, since eigenvectors are defined as non-zero vectors, every non-zero vector in $ \mathbb{R}^n $ is an eigenvector corresponding to the eigenvalue $ \lambda = 1 $. 3. Existence of an Eigenbasis: An eigenbasis is a basis for the vector space consisting entirely of eigenvectors of the matrix. Since every non-zero vector is an eigenvector of $ I_n $, any basis of $ \mathbb{R}^n $ is an eigenbasis for $ I_n $. For example, consider the standard basis vectors in $ \mathbb{R}^n $: \[ \mathbf{e}_1 = \begin{pmatrix}1 \\ 0 \\ \vdots \\ 0\end{pmatrix}, \quad \mathbf{e}_2 = \begin{pmatrix}0 \\ 1 \\ \vdots \\ 0\end{pmatrix}, \quad \dots, \quad \mathbf{e}_n = \begin{pmatrix}0 \\ 0 \\ \vdots \\ 1\end{pmatrix} \] Since $ I_n \mathbf{e}_i = \mathbf{e}_i $ for each $ i = 1, 2, \dots, n $, the set $ \{ \mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n \} $ forms an eigenbasis for $ I_n $. ### Conclusion: Yes, there exists an eigenbasis for the identity matrix $ I_n $. In fact, any basis of the vector space $ \mathbb{R}^n $ (or $ \mathbb{C}^n $) serves as an eigenbasis for $ I_n $, since all non-zero vectors are eigenvectors corresponding to the eigenvalue $ \lambda = 1 $.	True
GPT-4o	Yes, there is an eigenbasis for the identity matrix $ I_n $. To understand why, let's break it down step-by-step: 1. Eigenvalues of the Identity Matrix: The identity matrix $ I_n $ is a square matrix where all the diagonal elements are 1, and all off-diagonal elements are 0. The eigenvalue equation for a matrix $ A $ is given by: \[ A \mathbf{v} = \lambda \mathbf{v} \] For the identity matrix $ I_n $, this becomes: \[ I_n \mathbf{v} = \lambda \mathbf{v} \] Since $ I_n \mathbf{v} = \mathbf{v} $ for any vector $ \mathbf{v} $, it follows that: \[ \mathbf{v} = \lambda \mathbf{v} \] This implies that $ \lambda $ must be 1. Thus, the only eigenvalue of the identity matrix $ I_n $ is $ \lambda = 1 $. 2. Eigenvectors of the Identity Matrix: Given that the eigenvalue $ \lambda = 1 $, the eigenvector equation becomes: \[ I_n \mathbf{v} = 1 \cdot \mathbf{v} \] This equation is satisfied by any non-zero vector $ \mathbf{v} $ in $ \mathbb{R}^n $. Therefore, every non-zero vector in $ \mathbb{R}^n $ is an eigenvector of $ I_n $ corresponding to the eigenvalue 1. 3. Eigenbasis: An eigenbasis for a matrix is a basis for the vector space consisting of eigenvectors of the matrix. Since every non-zero vector in $ \mathbb{R}^n $ is an eigenvector of $ I_n $ with eigenvalue 1, we can choose the standard basis vectors $ \mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n $ of $ \mathbb{R}^n $ as our eigenbasis. These standard basis vectors are: \[ \mathbf{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \quad \mathbf{e}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \quad \ldots, \quad \mathbf{e}_n = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix} \] Each of these vectors satisfies: \[ I_n \mathbf{e}_i = \mathbf{e}_i \] for $ i = 1, 2, \ldots, n $, confirming that they are eigenvectors corresponding to the eigenvalue 1. Therefore, the set $ \{ \mathbf{e}_1, \mathbf{e}_2, \ldots, \mathbf{e}_n \} $ forms an eigenbasis for the identity matrix $ I_n $.	True
llama3.1-70B-Instruct	Yes, there is an eigenbasis for the identity matrix I_n. The identity matrix I_n is a diagonal matrix with all diagonal entries equal to 1. Therefore, all the eigenvalues of I_n are 1. To find the eigenvectors of I_n, we need to solve the equation: I_n v = λ v where v is an eigenvector and λ is an eigenvalue. Since all the eigenvalues of I_n are 1, we can write: I_n v = v This equation is satisfied for any vector v, since the identity matrix I_n does not change the vector v. Therefore, every vector v is an eigenvector of I_n, and the set of all vectors v forms an eigenbasis for I_n. In other words, the identity matrix I_n has an eigenbasis consisting of all vectors in the vector space, and the corresponding eigenvalues are all equal to 1.	True
llama3.1-405B-Instruct	Yes, there is an eigenbasis for the identity matrix I_n. The identity matrix I_n has n eigenvalues, all of which are equal to 1. The corresponding eigenvectors are the standard basis vectors of R^n, which are: e_1 = (1, 0, 0, ..., 0) e_2 = (0, 1, 0, ..., 0) ... e_n = (0, 0, 0, ..., 1) These eigenvectors are linearly independent and span R^n, so they form a basis for R^n. Therefore, the identity matrix I_n has an eigenbasis consisting of the standard basis vectors of R^n.	True
DeepSeek-Coder-V2	Yes, there is an eigenbasis for the identity matrix $ I_n $. To understand this, let's first recall some basic definitions and properties: 1. Identity Matrix $ I_n $: This is an $ n \times n $ matrix with ones on the diagonal and zeros elsewhere. It is denoted as $ I_n $. 2. Eigenvalue and Eigenvector: For a matrix $ A $, a scalar $ \lambda $ is called an eigenvalue and a non-zero vector $ v $ is called an eigenvector if they satisfy the equation $ Av = \lambda v $. 3. Eigenbasis: An eigenbasis is a basis for the vector space consisting entirely of eigenvectors. Now, let's analyze the identity matrix $ I_n $: - The identity matrix $ I_n $ has all its diagonal entries as 1, and all off-diagonal entries as 0. - For any vector $ v $ in $ \mathbb{R}^n $ (or $ \mathbb{C}^n $), multiplying $ v $ by $ I_n $ results in $ I_n v = v $. This implies that every non-zero vector $ v $ in $ \mathbb{R}^n $ (or $ \mathbb{C}^n $) is an eigenvector of $ I_n $ with eigenvalue 1. Specifically: \[ I_n v = 1 \cdot v \] Since every non-zero vector in $ \mathbb{R}^n $ (or $ \mathbb{C}^n $) is an eigenvector of $ I_n $ with eigenvalue 1, we can choose any basis for $ \mathbb{R}^n $ (or $ \mathbb{C}^n $), and this basis will consist entirely of eigenvectors of $ I_n $. Therefore, there exists an eigenbasis for the identity matrix $ I_n $. For example, the standard basis vectors $ \{e_1, e_2, \ldots, e_n\} $ where $ e_i $ is the vector with 1 in the $ i $-th position and 0 elsewhere, form an eigenbasis for $ I_n $. In conclusion, the answer to the problem is yes, there is an eigenbasis for the identity matrix $ I_n $.	True
Qwen2-72B-Instruct	Yes, there is an eigenbasis for the identity matrix $I_n$. Explanation: An eigenbasis of a matrix is a basis for the vector space consisting of eigenvectors of that matrix. For a matrix to have an eigenbasis, it needs to be diagonalizable, meaning it can be expressed as $A = PDP^{-1}$, where $D$ is a diagonal matrix whose diagonal entries are the eigenvalues of $A$, and $P$ is the matrix whose columns are the eigenvectors of $A$. The identity matrix $I_n$ is a diagonal matrix with all diagonal entries equal to 1. Therefore, its eigenvalues are all 1, and any non-zero vector is an eigenvector of $I_n$ since $I_n v = v$ for any non-zero vector $v$. As a result, any basis of the vector space can be chosen as an eigenbasis for $I_n$, since all vectors in the basis will be eigenvectors of $I_n$. In summary, any basis of the vector space $\mathbb{R}^n$ (or whatever the underlying vector space is) can serve as an eigenbasis for the identity matrix $I_n$.	True
baichuan4	Yes, there is an eigenbasis for the identity matrix $I_n$. An eigenbasis of a matrix is a basis consisting of the eigenvectors of the matrix. For the identity matrix $I_n$, every non-zero vector is an eigenvector with the corresponding eigenvalue of 1. This is because multiplying any vector by the identity matrix results in the same vector: $$I_n \mathbf{v} = \mathbf{v}$$ Therefore, any set of $n$ linearly independent vectors in $\mathbb{R}^n$ (or $\mathbb{C}^n$) can serve as an eigenbasis for $I_n$. A standard choice for this basis is the set of standard basis vectors $e_1, e_2, \ldots, e_n$, where $e_i$ is the vector with a 1 in the $i$-th position and 0s elsewhere. However, any other set of linearly independent vectors would also form an eigenbasis for $I_n$.	True
doubao	The identity matrix I_n has n linearly independent eigenvectors, which can form an eigenbasis. The eigenvectors of I_n are the standard basis vectors e_i (where i = 1, 2,..., n), and the corresponding eigenvalue for each eigenvector is 1. So, yes, there is an eigenbasis for the identity matrix I_n.	True
minimax	An eigenbasis for a matrix is a set of eigenvectors that form a basis for the vector space in which the matrix operates. The identity matrix $ I_n $ is an $ n \times n $ matrix with ones on the diagonal and zeros elsewhere. Mathematically, it is defined as: \[ I_n = \begin{bmatrix} 1 & 0 & \cdots & 0 \\ 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 1 \end{bmatrix} \] The eigenvalues of the identity matrix $ I_n $ are all equal to 1. This is because if you multiply $ I_n $ by any vector $ \mathbf{v} $, you get $ I_n \mathbf{v} = \mathbf{v} $, which means that the vector $ \mathbf{v} $ is an eigenvector corresponding to the eigenvalue $ \lambda = 1 $. Since the identity matrix is diagonal, its eigenvectors are simply the standard basis vectors of the space. For $ \mathbb{R}^n $, these are: \[ \mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{bmatrix}, \mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{bmatrix}, \ldots, \mathbf{e}_n = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{bmatrix} \] Each of these vectors is an eigenvector of $ I_n $ corresponding to the eigenvalue 1. Since there are $ n $ such eigenvectors and they are linearly independent, they form a basis for $ \mathbb{R}^n $. Therefore, the standard basis is an eigenbasis for the identity matrix $ I_n $. In conclusion, yes, there is an eigenbasis for the identity matrix $ I_n $, and it is the standard basis of the vector space.	True
kimi	Yes, there is an eigenbasis for the identity matrix $ I_n $. Solution: 1. Eigenvalues of Identity Matrix: The identity matrix $ I_n $ is a square matrix of size $ n \times n $ with ones on the diagonal and zeros elsewhere. For any vector $ v $ that is not the zero vector, the product $ I_n \cdot v $ will result in the same vector $ v $. This means that the identity matrix acts as the multiplicative identity for vectors, just like the number 1 acts as the multiplicative identity for numbers. 2. Eigenvectors: For a matrix $ A $, a vector $ v $ is an eigenvector if there exists a scalar $ \lambda $ such that $ A \cdot v = \lambda \cdot v $. Here, $ \lambda $ is called the eigenvalue corresponding to the eigenvector $ v $. 3. Eigenbasis for $ I_n $: For the identity matrix $ I_n $, every non-zero vector $ v $ in $ \mathbb{R}^n $ is an eigenvector with the eigenvalue $ \lambda = 1 $. This is because $ I_n \cdot v = v $, which satisfies the eigenvector equation with $ \lambda = 1 $. 4. Constructing the Eigenbasis: The standard basis vectors of $ \mathbb{R}^n $, which are the vectors $ e_1, e_2, \ldots, e_n $ where each $ e_i $ has a 1 in the $ i $-th position and zeros elsewhere, are all eigenvectors of $ I_n $ with eigenvalue 1. 5. Conclusion: Since every standard basis vector is an eigenvector of $ I_n $ with eigenvalue 1, and these vectors form a basis for $ \mathbb{R}^n $, they constitute an eigenbasis for $ I_n $. This eigenbasis is orthogonal and spans the entire space $ \mathbb{R}^n $. Thus, the identity matrix $ I_n $ has an eigenbasis consisting of its standard basis vectors, each with eigenvalue 1.	True
wenxinyiyan	Yes, there is an eigenbasis for the identity matrix I_n. To see this, consider any nonzero vector v in R^n. We have: I_n * v = v This means that v is an eigenvector of I_n with eigenvalue 1. Since we can choose any nonzero vector in R^n to be an eigenvector, and these vectors will span R^n, we can construct an eigenbasis for I_n. Specifically, let {v_1, v_2, ..., v_n} be any basis for R^n (for example, the standard basis). Then each v_i is an eigenvector of I_n with eigenvalue 1, and {v_1, v_2, ..., v_n} is an eigenbasis for I_n.	True
zhipu	Yes, there is an eigenbasis for the identity matrix $ I_n $. To understand why, let's first recall some definitions: 1. Eigenvalue and Eigenvector: For a square matrix $ A $, a non-zero vector $ v $ is called an eigenvector if there exists a scalar $ \lambda $ (called the eigenvalue) such that $ A v = \lambda v $. 2. Eigenbasis: A set of eigenvectors of a matrix that forms a basis for the vector space. Now, let's consider the identity matrix $ I_n $, which is an $ n \times n $ matrix with 1's on the diagonal and 0's elsewhere. ### Eigenvalues of $ I_n $ For the identity matrix $ I_n $, any vector $ v $ satisfies: \[ I_n v = v \] This implies that every non-zero vector $ v $ is an eigenvector corresponding to the eigenvalue $ \lambda = 1 $. Therefore, the only eigenvalue of $ I_n $ is 1, and it has multiplicity $ n $. ### Eigenvectors of $ I_n $ Since every non-zero vector $ v $ in $ \mathbb{R}^n $ satisfies $ I_n v = v $, every non-zero vector in $ \mathbb{R}^n $ is an eigenvector corresponding to the eigenvalue 1. ### Eigenbasis for $ I_n $ To form an eigenbasis, we need $ n $ linearly independent eigenvectors. In the case of $ I_n $, we can choose any set of $ n $ linearly independent vectors in $ \mathbb{R}^n $. A common choice is the standard basis $ \{e_1, e_2, \ldots, e_n\} $, where: \[ e_1 = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \quad e_2 = \begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix}, \quad \ldots, \quad e_n = \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix} \] These vectors are linearly independent and span $ \mathbb{R}^n $, thus forming a basis. Since each $ e_i $ is an eigenvector of $ I_n $ corresponding to the eigenvalue 1, $ \{e_1, e_2, \ldots, e_n\} $ is an eigenbasis for $ I_n $. ### Conclusion Therefore, there is indeed an eigenbasis for the identity matrix $ I_n $, and any set of $ n $ linearly independent vectors in $ \mathbb{R}^n $ can serve as such an eigenbasis. The standard basis $ \{e_1, e_2, \ldots, e_n\} $ is a straightforward example of such an eigenbasis.	True
gpto1	Yes, there is an eigenbasis for the identity matrix $ I_n $. In fact, every non-zero vector in $ \mathbb{R}^n $ (or $ \mathbb{C}^n $) is an eigenvector of the identity matrix, corresponding to the eigenvalue $ \lambda = 1 $. ### Explanation: The identity matrix $ I_n $ is a square $ n \times n $ matrix with ones on the main diagonal and zeros elsewhere: \[ I_n = \begin{pmatrix} 1 & 0 & 0 & \dots & 0 \\ 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \dots & 1 \\ \end{pmatrix} \] An eigenvector $ \mathbf{v} $ of a matrix $ A $ is a non-zero vector that satisfies the equation: \[ A\mathbf{v} = \lambda \mathbf{v} \] where $ \lambda $ is the eigenvalue corresponding to $ \mathbf{v} $. #### For the Identity Matrix: 1. Eigenvalue Determination: Let's find the eigenvalues by solving the characteristic equation: \[ \det(I_n - \lambda I_n) = 0 \] Simplify $ I_n - \lambda I_n = (1 - \lambda)I_n $. The determinant of $ (1 - \lambda)I_n $ is: \[ \det((1 - \lambda)I_n) = (1 - \lambda)^n \det(I_n) = (1 - \lambda)^n (1) = (1 - \lambda)^n \] Setting the determinant to zero: \[ (1 - \lambda)^n = 0 \implies \lambda = 1 \] So, the only eigenvalue of $ I_n $ is $ \lambda = 1 $, and its algebraic multiplicity is $ n $. 2. Eigenvector Determination: For $ \lambda = 1 $, the eigenvector equation becomes: \[ I_n \mathbf{v} = \mathbf{v} \] Which simplifies to: \[ \mathbf{v} = \mathbf{v} \] This equation is always true for any vector $ \mathbf{v} $ in $ \mathbb{R}^n $ (or $ \mathbb{C}^n $). However, since eigenvectors are defined as non-zero vectors, every non-zero vector in $ \mathbb{R}^n $ is an eigenvector corresponding to the eigenvalue $ \lambda = 1 $. 3. Existence of an Eigenbasis: An eigenbasis is a basis for the vector space consisting entirely of eigenvectors of the matrix. Since every non-zero vector is an eigenvector of $ I_n $, any basis of $ \mathbb{R}^n $ is an eigenbasis for $ I_n $. For example, consider the standard basis vectors in $ \mathbb{R}^n $: \[ \mathbf{e}_1 = \begin{pmatrix}1 \\ 0 \\ \vdots \\ 0\end{pmatrix}, \quad \mathbf{e}_2 = \begin{pmatrix}0 \\ 1 \\ \vdots \\ 0\end{pmatrix}, \quad \dots, \quad \mathbf{e}_n = \begin{pmatrix}0 \\ 0 \\ \vdots \\ 1\end{pmatrix} \] Since $ I_n \mathbf{e}_i = \mathbf{e}_i $ for each $ i = 1, 2, \dots, n $, the set $ \{ \mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n \} $ forms an eigenbasis for $ I_n $. ### Conclusion: Yes, there exists an eigenbasis for the identity matrix $ I_n $. In fact, any basis of the vector space $ \mathbb{R}^n $ (or $ \mathbb{C}^n $) serves as an eigenbasis for $ I_n $, since all non-zero vectors are eigenvectors corresponding to the eigenvalue $ \lambda = 1 $.	True
deepseekv3	Yes, there is an eigenbasis for the identity matrix $ I_n $. Here's why: 1. Eigenvalues of $ I_n $: The identity matrix $ I_n $ has only one eigenvalue: $ \lambda = 1 $. This is because for any vector $ \mathbf{v} $, $ I_n \mathbf{v} = \mathbf{v} $. 2. Eigenspace: The eigenspace corresponding to $ \lambda = 1 $ is the entire vector space $ \mathbb{R}^n $ (or $ \mathbb{C}^n $, depending on the field), since every vector is an eigenvector of $ I_n $ with eigenvalue $ 1 $. 3. Eigenbasis: Any basis of $ \mathbb{R}^n $ (or $ \mathbb{C}^n $) is an eigenbasis for $ I_n $. For example, the standard basis $ \{ \mathbf{e}_1, \mathbf{e}_2, \dots, \mathbf{e}_n \} $ is an eigenbasis for $ I_n $, where $ \mathbf{e}_i $ is the vector with a $ 1 $ in the $ i $-th position and $ 0 $ elsewhere. Thus, $ I_n $ has an eigenbasis, and in fact, any basis of the vector space serves as an eigenbasis for $ I_n $.	True