Solution.

Consider the following equation. \[ T(0, 1) = (0, -1) = -1 \cdot (0, 1) \] Therefore, \((0, 1)\) is an eigenvector of \(T\) and \(-1\) is the corresponding eigenvalue of \(T\).

Solution.

For some scalar \(\lambda \), consider the following equation. \[ T(a, b) = (b, -a) = \lambda (a, b) = (\lambda a, \lambda b) \] In other words \(b = \lambda a\) and \(-a = \lambda b\). Substituting, \(-a = \lambda ^2 a\) and \((\lambda ^2 + 1) a = 0\). Because \(\lambda \in \mathbb {R}\), \(a = 0\) and \(b = 0\). However, because eigenvectors are nonzero vectors by definition, \(T\) has not real eigenvalues and eigenvectors.

Proof.

First, let \(B = \{ v_1, \ldots , v_n \}\) be a basis for \(V\) and let the matrix representation of \(T\) be a diagonal matrix \(A\) whose diagonal entries are \(\lambda _1, \ldots , \lambda _n\). Thus \[ (Tv_j)_B = \begin {bmatrix} 0 \\ \vdots \\ \lambda _j \\ \vdots \\ 0 \end {bmatrix} \] hold by construction and \(Tv_j = \lambda _j v_j\). In other words, \(v_j\) and \(\lambda _j\) are an eigenvector and an eigenvalue of \(T\) and there exists a basis composed of only eigenvectors of \(T\).

Continuing, let \(B = \{ v_1, \ldots , v_n \}\) be a basis for \(V\) whose elements are all eigenvectors of \(T\). Therefore, \(Tv_j = \lambda v_j\) and the following equation holds. \[ (Tv_j)_B = \begin {bmatrix} 0 \\ \vdots \\ \lambda _j \\ \vdots \\ 0 \end {bmatrix} \] Therefore, there exists a basis whose matrix representation is a diagonal matrix.

Solution.

Consider the following equation. \[ \begin {bmatrix} 1 & 0 \\ 0 & -1 \end {bmatrix} \begin {bmatrix} 0 \\ 1 \end {bmatrix} = \begin {bmatrix} 0 \\ -1 \end {bmatrix} = -1 \cdot \begin {bmatrix} 0 \\ 1 \end {bmatrix} \] Thus, \(-1\) is an eigenvalue and \([0 1]\) is an eigenvector.

Proof.

Notice that \(Tv = \lambda v\) if and only if \(A_B^B (v)_B = (Tv)_B = (\lambda v)_B = \lambda (v)_B\). Thus, the theorem holds.

Proof.

First, notice that because \(S_\lambda \subset V\), it suffices to show that the set is closed under vector addition and scalar multiplication. Consider the following equations for \(u, v \in S_\lambda \) and some scalars \(\alpha \) and \(\beta \). \begin{align*} T(\alpha u + \beta v) &= \alpha Tu + \beta Tv = \alpha \lambda u + \beta \lambda v \\ &= \lambda (\alpha u + \beta v) \end{align*}

In other words, \(\alpha u + \beta v \in S_\lambda \) and the theorem holds.

Proof.

First notice that by definition, \(\lambda \) is an eigenvalue of \(A\) if and only if \(Ax = \lambda x\) holds for some nonzero vector \(x\). Continuing, \(Ax = \lambda x\) holds if and only if \(Ax = \lambda I_n x\) and \((A - \lambda I_n)x = 0\). In other words, the columns of the matrix \(A - \lambda I_n\) are linearly dependent and invertible. Therefore by Theorem ??, the theorem holds.

Solution.

To find all eigenvalues of \(A\), it suffices to find all solutions for the characteristic equation of \(A\) by Theorem 0.1.10 . \[ A - \lambda I_2 = \begin {bmatrix} 1 - \lambda & 0 \\ 0 & -1 - \lambda \end {bmatrix} \] Therefore, \(\det (A - \lambda I_2) = (1 - \lambda )(-1 - \lambda )\) and \(\lambda = 1\) or \(\lambda = -1\) are the solutions to the characteristic equation.

Continuing, the corresponding eigenspaces could be found. For \(\lambda = 1\), the eigenspace is defined as \(S_1 = \{ x | Ax = x \}\). Solving for the equation \((A - I)x = 0\), the following equations hold. \begin{align*} \begin {bmatrix} 0 & 0 \\ 0 & -2 \end {bmatrix} \begin {bmatrix} x_1 \\ x_2 \end {bmatrix} &= 0 \\ -2x_2 &= 0 \end{align*}

Therefore, the eigenspace for the eigenvalue \(1\) can be represented as the following \[ S_1 = \Span \left \{ \begin {bmatrix} 1 \\ 0 \end {bmatrix} \right \} \] Similarly, \(S_{-1} = \{ x | Ax = -x \}\) and \((A + I)x = 0\) could be solved. \begin{align*} \begin {bmatrix} 2 & 0 \\ 0 & 0 \end {bmatrix} \begin {bmatrix} x_1 \\ x_2 \end {bmatrix} &= 0 \\ 2x_1 &= 0 \end{align*}

Therefore, \[ S_{-1} = \Span \left \{ \begin {bmatrix} 0 \\ 1 \end {bmatrix} \right \} \] and the corresponding eigenspaces for all eigenvalues of \(A\) are found.

Proof.

By definition of similar matrices, there exists some invertible matrix \(P\) such that \(B = P^{-1} A P\). Therefore, the following equations hold. \[ P^{-1} (A - \lambda I) P = P^{-1} A P - P^{-1} \lambda I P = B - \lambda I \] Therefore, \(\det (B - \lambda I) = \det (P^{-1} (A - \lambda I) P) = \det (A - \lambda I)\) and the characteristic equations are the same. In other words, the eigenvalues for \(A\) and \(B\) are equal and the theorem holds.

Proof.

By definition, notice that for a triangular matrix \(A\), \(A - \lambda I\) is also triangular. Therefore by Theorem ??, \[ \det (A - \lambda I) = \prod _{k=1}^n (\lambda _k - \lambda ) \] and \(\det (A - \lambda I_n) = 0\) if and only if \(\lambda = \lambda _k\). Because \(\lambda _k\) are the diagonal elements of \(A\), the theorem holds.

Proof.

A square matrix \(A\) contains a zero eigenvalue if and only if \(\det (A - 0 I) = 0\) and \(\det (A) = 0\). Because \(A\) is invertible if and only if \(\det (A) \neq 0\) by Theorem ??, the theorem holds.

Proof.

By definition, \(Ax = \lambda x\) and \(x = A^{-1} \lambda x\). Continuing, \(A^{-1} x = \frac {1}{\lambda } \cdot x\). Therefore, \(x\) is also an eigenvector of \(A^{-1}\) with the corresponding eigenvalue \(\frac {1}{\lambda }\).

Proof.

First, notice that \(\{ x_1 \}\) is linearly independent. Therefore, it suffices to show that if the theorem holds for \(k - 1 > 0\), then the statement holds for \(k\). Assume \(\{ x_1, \ldots , x_{k-1} \}\) is a linearly independent set of eigenvectors with distinct corresponding eigenvalues \(\lambda _1, \ldots , \lambda _{k-1}\). Consider the following equations for some scalars \(c_i\). \begin{align*} \sum _{i=1}^k c_i x_i &= 0 \\ \sum _{i=1}^k c_i A x_i &= 0 \\ \sum _{i=1}^k c_i \lambda _i x_i &= 0 \\ \sum _{i=1}^{k-1} c_i \lambda _k x_i &= 0 \end{align*}

Subtracting the last two equations, \(\sum _{i=1}^{k-1} c_i (\lambda _i - \lambda _k) x_i = 0\) is obtained. Because \(\{ x_1, \ldots , x_{k-1} \}\) is linearly independent, \(c_i (\lambda _i - \lambda _k) = 0\) for all integer \(i \in [1, k)\). Continuing, because \(\lambda _i - \lambda _k \neq 0\) by construction, \(c_i = 0\) for all integer \(i \in [1, k)\) and \(c_k = 0\) since \(x_k \neq 0\). Thus \(\{ x_1, \ldots , x_k \}\) is linearly independent and the theorem holds by induction.

Proof.

By definition, the eigenspace for \(\lambda \) is the set of all vectors \(v\) that satisfy \(Av = \lambda v\). Because \((A - \lambda ) v = 0\), \(S_\lambda \) can be interpreted as \(\ker (A - \lambda I)\). Moreover by Theorem ??, the following equation holds. \[ \Rank (A - \lambda I) + \Null (A - \lambda I) = n \] Because \(\dim (\ker (A - \lambda I)) = \Null (A - \lambda I)\) holds by definition, \(\dim (S_\lambda ) = n - \Rank (A - \lambda I)\).

Proof.

Consider the following equation. \[ (kA)x = k(Ax) = k \lambda x = (k \lambda ) x \] Therefore, \(x\) is an eigenvector of \(kA\) and its corresponding eigenvalue is \(k \lambda \).

Proof.

Assume that if the statement holds for \(n - 1\), then the statement holds for \(n > 1\). Consider the following equations. \begin{align*} A^n x &= A \left ( A^{n-1} x \right ) = A \left ( \lambda ^{n-1} x \right ) = \lambda ^{n-1} Ax \\ &= \lambda ^{n-1} \lambda x = \lambda ^n x \end{align*}

Because \(Ax = \lambda x\), the statement holds by induction.

Proof.

By the Fundamental Theorem of Algebra, the following equation holds. \[ \det (A - \lambda I) = \prod _{k=1}^n (\lambda _k - \lambda ) = 0 \] Therefore, all eigenvalues are the diagonal elements and by the definition of trace of \(A\), the sum of all eigenvalues are the trace of \(A\).

Proof.

Substituting \(\lambda = 0\) into the equation \[ \det (A - \lambda I) = \prod _{k=1}^n (\lambda _k - \lambda ), \] \(\det (A) = \prod _{k=1}^n \lambda _k\) and the equation holds.

Proof.

First and foremost, notice that \(A\) and a diagonal matrix \(D\) of the same order with entries \(\lambda _i\) are similar if and only if there exists an invertible matrix \(P\) such that \(P^{-1} A P = D\) or \(AP = PD\). Let \(x_j\) and \(e_j\) be the \(j^\text {th}\) column of \(P\) and \(I\) respectively. Then, the following equations hold. \begin{align*} A x_j &= A Pe_j = PDe_j = P \lambda _j e_j \\ &= \lambda _j P e_j = \lambda _j x_j \end{align*}

Thus, \(x_j\) and \(\lambda _j\) are an eigenvector and the corresponding eigenvalue of \(A\). By definition, because \(M\) is invertible, \(x_j\) are all linearly independent and \(A\) has \(n\) linearly independent eigenvectors. In other words, if \(A\) is diagonalizable, then \(A\) has \(n\) linearly independent eigenvectors.

Continuing, the converse can be proven. Let \(A\) have \(n\) linearly independent eigenvectors denoted as \(x_j\) with corresponding eigenvalues \(\lambda _j\). Construct matrices \(P\) and \(D\) such that the \(j^\text {th}\) column of \(P\) is \(x_j\) and the diagonal element of \(D\) is \(\lambda _j\). If \(e_j\) is the \(j^\text {th}\) column of \(I\), then the \(j^\text {th}\) column of \(AP\) can be written as the following. \[ APe_j = A x_j = \lambda _j x_j \] Similarly, the \(j^\text {th}\) column of \(PD\) can be written as \(PDe_j = M \lambda _j e_j = \lambda _j Me_j = \lambda _j x_j\) and \(AP = PD\) holds. Moreover, because \(P\) is invertible by construction, \(P^{-1} A P = D\) and \(A\) is diagonalizable.

Proof.

First and foremost, notice that by Theorem 0.1.17 , the set of corresponding eigenvectors \(\{ x_1, \ldots , x_n \}\) of \(A\) is linearly independent. Thus by Theorem 0.1.23 , the corollary holds.

Proof.

First and foremost, notice that by Theorem 0.1.13 , because \(A\) and \(D_1\) are similar, they have the same characteristic equation. Similarly, because \(A\) and \(D_2\) are similar, they also have the same characteristic equation and the characteristic equations of \(D_1\) and \(D_2\) are equal. \[ \det (D_1 - \lambda I) = \det (D_2 - \lambda I) = 0 \] By construction, notice that \(D_1 - \lambda I\) and \(D_2 - \lambda I\) are diagonal matrices. Thus by Corollary ??, the determinant is the product of all \((\lambda - \lambda _k\) for integer \(k \in [1, n]\) where \(\lambda _k\) are the diagonal entries for \(D_1\). Because multiplication is commutative, the diagonal entries for \(D_1\) and \(D_2\) are the same.

Proof.

First, the statement if \(\sum _{i=1}^k \dim (S_{\lambda _i}) = n\), then \(A\) is diagonalizable can be proven. Let \(B_i\) be a basis for eigenspace \(S_{\lambda _i}\). By definition, vectors \(v\) in eigenspace \(S_{\lambda _i}\) satisfy \(Av = \lambda _i v\). Because \(\lambda _i\) are all distinct, no eigenspace shares the same eigenvector and thus no basis of two different eigenspaces shares a vector. In other words, \(B\) contains \(n\) distinct eigenvectors of \(A\). Let \(v_i \in S_{\lambda _i}\) be some linear combination of vectors in \(B_i\). Notice that if \(\sum _{i=1}^k v_i = 0\), then the coefficients must be zero since \(\{ v_1, \ldots , v_k \}\) is linearly independent by Theorem 0.1.17 . In other words, all coefficients for the linear combination of \(B_i\) that forms \(v_i\) must be zero for \(\sum _{i=1}^k v_i = 0\) to hold. Thus, \(B\) is linearly independent and \(A\) is diagonalizable by Theorem 0.1.23 .

Continuing, the converse can be proven. Assume that \(\sum _{i=1}^k \dim (S_{\lambda _i}) > n\). Continuing from the previous definition, \(B \subset \mathbb {R}^n\) and the inequality \(\sum _{i=1}^k \dim (S_{\lambda _i}) > n\) leads to contradiction. Moreover, if \(\sum _{i=1}^k \dim (S_{\lambda _i}) < n\) then \(A\) has no set that contains \(n\) linearly independent eigenvectors. Thus, \(A\) is not diagonalizable if \(\sum _{i=1}^k \dim (S_{\lambda _i}) \neq n\).