My question is about linear algebra, especially invariant subspaces and  diagonalizability. Here is the question:  Let $A$ be a diagonalizable linear operator on the finite dimensional vector space over a field, and let $W$ be a subspace of $V$  which is invariant under $A$. Let us denote by $A_W$ the restriction of $A$ to $W$. I need to prove that $A_W$ is diagonalizable.  How can I prove that?  thanks for your help...
85147
Suppose $V$ is a vector space over $\mathbb{C}$, and $A$ is a linear transformation on $V$ which is diagonalizable. I.e. there is a basis of $V$ consisting of eigenvectors of $A$. If $W\subseteq V$ is an invariant subspace of $A$ (so $A(W)\subseteq W$), show that $A|_W$ is also diagonalizable.   I tried supposing $A$ has distinct eigenvalues $\lambda_1,\ldots,\lambda_m$, with $V_i=\{v\in V: Av=\lambda_i v\}$. Then we can write $V=V_1\oplus\cdots\oplus V_m,$ but I'm not sure whether it is true that  $$W=(W\cap V_1)\oplus\cdots\oplus (W\cap V_m),.$$  If it is true, then we're done, but it may be wrong.
29645
Consider the linear dynamical system $\dot{x} = Ax $ in $V$ a finite dimensional vector space.  The definition of an invariant subspace $U$ is as follows:  For all $x_0 \in U$, (initial condition), the solution $x(t)$ with initial condition $x(0)=x_0$ lies in $U$, i.e. $x(t) \in U$ for all $t \in \mathbb{R}$. Intuitively, it means if we start in a subspace we never leave that subspace.  Prove that a subspace $U \subset V$ is an invariant subspace if and only if $AU \subset U$.
18127
Show that every idempotent matrix is diagonalizable. What can you say if $A$ is tripotent ($A^3=A?$) What if $A^k=A?$  The first two cases is obvious since we can find the minimal polynomial to be $t(t-1)$ or $t(t-1)(t+1)$, which are products of distinct linear factors. However, what can we say about the general case $A^k=A?$ Any solutions, hints, or suggestions would be appreciated.
91187
Let $V$ be a finite-dimensional $K$-vector space and let $T$ be a linear operator from $V$ to $V$.  I already proved that for every polynomial $p(x) \in K[x]$, $\ker p(T)$ and $Im p(T)$ are $T$-invariant. Is it true that for every $W$ proper $T$-invariant subspace, $W$ is the kernel of $p(T)$ for a polynomial $p$?  My guess is yes. If $p$ is the characteristic polynomial of the restriction of $T$ to $W$, then $p(T)v=0$ for every $v$ and then $W\subset \ker p(T)$. How can I finish the converse?
117274
I'm having troble proving this: Let $A\in M_n(R). $ If $n=2$ and $A$ has exactly one eigenvalue $\lambda$ then $A$ is similar to $\begin{bmatrix}  \lambda &amp; 1\\ 0 &amp; \lambda \\ \end{bmatrix}$ or to $\begin{bmatrix}  \lambda &amp; 0\\ 0 &amp; \lambda \\ \end{bmatrix}$ I did prove that if the geometric multiplicity of $\lambda$ is $2$, then $A$ is similar to $\begin{bmatrix}  \lambda &amp; 0\\ 0 &amp; \lambda \\ \end{bmatrix}$, but I'm struggling with the case when $A$ is not diagonalizable. Any thoughts?
87552
Let $A$ be a diagonalizable matrix and $\lambda$ be an eigenvalue of A. Prove that rank($\lambda I - A$) = rank(($\lambda I - A)^2$).  The hint they provided is that for any invertible matrix B, rank(AB) = rank (A) and rank(BA) = rank(A).
98701
If $A$ is a $8\times 8$ matrix over $\mathbb{R}$ and $A^3=A$ then prove that $A$ is diagonalizable.   I have got that the minimal polynomial of $A$ may be $x^3-x$ or $x$ or $x(x+1)$ or $x(x-1)$ in the 2nd case it is not possible in the 3rd and 4th cases the matrix will be the identity matrix or identity matrix multiplied by a scalar!  But I cannot reach any further.
110656
Let $A$ $\in M_3(\mathbb R)$ which is not a diagonal matrix. Pick out the cases when $A$ is diagonalizable over $\mathbb R$:  a. when $A^2 = A$;  b. when $(A - 3I)^2 = 0$;  c. when $A^2 + I = 0$.  I could eliminate c. by using the equation $\lambda^2+1$ and showing if a matrix has to satisfy this then it has to be diagonal.  I am also in doubt whether all the eigen values of $A$ satisfy this or not.
43037
This is a question from Finite-Dimensional Linear Algebra by Mark S. Gockenbach page 57 (exercise 2.5.15). I hope to check my proof.  Let $V$ be a vector space over a field $F$, and suppose $S$ and $T$ are subspaces of $V$ satisfying $S \cap T = \{0\}$. Suppose $\{s_1,s_2, ..., s_k\} \subset S$ and $\{t_1,t_2, ..., t_l\} \subset T$ are both linearly independent sets. Prove that $\{s_1,s_2, ..., s_k,t_1,t_2, ..., t_l \}$ is a linearly independent subset of $V$.  My proof) Suppose that $\alpha_1 s_1 +...+ \alpha_k s_k + \alpha_{k+1}t_1 +...+ \alpha_{k+l}t_l = 0$ for some $\{ \alpha_i\}_{i=1}^{k+l} \subset F$ Then $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_l = -(\alpha_1 s_1 + ... + \alpha_k s_k) \in S$ Since $\{ t_1,...t_l \} \subset T$ and $T$ is a subspace of $V$, $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_1 \in T $. Since $S \cap T = \{0 \}$, $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_l = 0$. Since $\{t_1, ... , t_l \}$ is a linearly independent set, $\alpha_{k+1} = ... = \alpha_{k+l} = 0$. By doing similarly, we can show that $\alpha_1 = ... = \alpha_k = 0$. Therefore, $\{s_1,s_2, ..., s_k,t_1,t_2, ..., t_l \}$ is a linearly independent subset of $V$. $\blacksquare$
26512
Let $A$ be an $n \times n$ matrix. Prove that if A is invertible, then there exists a polynomial $p$, such that $A^{-1}=p(A)$  Thus far:  Let $W$ denote the $k$ dimensional A-cyclic subspace spanned by a vector $v$.  Then, $I_n=\sum_{i=0}^{k} a_{i}A^i$ for some scalar $a_i$.  What I want to do is multiply both sides by $A^{-1}$, so that way it just falls into my hands, but Im not sure that is 'legal'.
86071
I am trying to prove that:  The matrix $C = \left(\begin{smallmatrix}A&amp; 0\\0 &amp; B\end{smallmatrix}\right)$ is diagonalizable, if only if $A$ and $B$ are diagonalizable.  If $A\in GL(\mathbb{C}^n)$ and $B\in GL(\mathbb{C}^m)$ are diagonalizable, then is easy to check the $C\in GL(\mathbb{C}^{n+m})$ is diagonalizable. But if I suppose that $C$ is diagonalizable, then exists $S = [S_1, S_2, \ldots, S_{n+m}]$, $S_i\in\mathbb{C}^{m+n}$, such that $S^{-1}CS = \mbox{diag}(\lambda_i)$ . Now $CS_i = \lambda_iS_i$, and  if $S_i = \left(\begin{smallmatrix}x_i\\y_i\end{smallmatrix}\right)$, $x_i\in\mathbb{C}^n$ and $y_i\in\mathbb{C}^m$, then $$Ax_i = \lambda_ix_i\quad\mbox{ and }\quad By_i = \lambda_iy_i.$$ So, if I can justify that $\{x_1,\ldots,x_{n+m}\}$ have exactly $n$ linear independent vectors and $\{y_1,\ldots,y_{n+m}\}$ have $m$ linear independent vectors, I will prove that $A$ and $B$ are diagonalizables, but I don't know how to prove that? Please, anyone have an idea? Thanks in advance.
97597
I want to prove that all orthogonal matrices are diagonalizable over $C$. I know that a matrix is orthogonal if $Q^TQ = QQ^T = I$ and $Q^T = Q^{-1}$, and that a matrix $A$ is diagonalizable if $A = PDP^{-1}$ where $D$ is a diagonal matrix. How can I start this proof?
63094
So i have this $3\times 3$ matrix  $$     \begin{pmatrix}     -2 &amp; 0 &amp; 0 \\     3 &amp; 1 &amp; -6 \\     0 &amp; 0 &amp; -2 \\     \end{pmatrix} $$  I want to check if the matrix is diagonizable. First thing i do is find the roots : namely $\lambda_1 = −2$ of multiplicity $2,$ and $\lambda_2 = 1$ of multiplicity $1.$  In order to ensure that the matrix is diagonalizable we need to find a basis of $\mathbb{R}^3$ whose elements are eigenvectors for $A.$ In particular, since we have the eigenvalue $\lambda_1=-2$ having multiplicity $2$ we need to check that the dimension of the eigenspace $E(\lambda_1)$ is equal to $2.$ We compute the eigenspaces of $A.$  $\DeclareMathOperator\Nul{Nul}$On the answer sheet is says that we need to find $\Nul(A+2Id)$... but why $2Id?$ Isn't is supposed to be $-2,$ so $\Nul(-2Id-A)$??   I do not understand this $\Nul(A+2Id).$ I always thought that we should compute $\Nul(-2Id-A),$ so what do i need to do ?   $\Nul(-2Id-A)$ gives me a different result from $\Nul(A+2Id).$  Does that mean that since $-2$ has a multiplicity of $2,$ we need to work with $2$ and not $-2$?  Somebody please help with a clear and simple explanation ! Thanks!
33091
If $V_0$ is the subspace consisting of matrices of the form $C=AB-BA$ for some $A,B$ in a vector space $V$ then $V_0=\{A\in V|\operatorname{Trace}(A)=0\}$.  The problem above is one of the past qualifying exam problems. I can prove that  $V_0\subset\{A\in V|\operatorname{Trace}(A)=0\}$ but I do not know what to do with converse.  Thank you in advance.
6236
Let A = $\begin{bmatrix}1 &amp; 1 &amp; 4\\0 &amp; 3 &amp; -4\\0&amp;0&amp;-1\end{bmatrix}$. Is the matrix A diagonalizable?  If so find a matrix P that diagonalizes A. Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors? Determine the eigendecomposition $A = PΛP^{-1}$. A = $\begin{bmatrix}3 &amp; 1 &amp; 0\\1 &amp; 2 &amp;1\\0&amp;1&amp;3\end{bmatrix}$. Is the matrix A diagonalizable?  If so find a matrix P that orthogonally diagonalizes A. Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors? (Note that A is real symmetric so that you do not have to compute the inverse of P).   My solution attempts:   det(λI - A) = $\begin{vmatrix}λ - 1 &amp; -1 &amp; -4\\0 &amp; λ - 3 &amp;4\\0&amp;0&amp;λ + 1\end{vmatrix}$ = (λ -1) (λ -3) (λ + 1) = 0   $λ_{1}$ = 1, $λ_{2}$ = 3, $λ_{3}$ = -1  ($λ_{1}I - A)x_{1}$ = 0  $\begin{bmatrix}0 &amp; -1 &amp; -4\\0 &amp; -2 &amp;4\\0&amp;0&amp;2\end{bmatrix}$ $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $\begin{bmatrix}0\\0 \\0\end{bmatrix}$   $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $x_{11} \begin{bmatrix}1\\0 \\0\end{bmatrix}$  It follows that $\begin{bmatrix}1\\0 \\0\end{bmatrix}$ is a basis for the eigenspace of A corresponding to $λ_{1}$ = 1.  In a similar way, I found other two basis vectors as $\begin{bmatrix}1/2\\1 \\0\end{bmatrix}$ and $\begin{bmatrix}-2/5\\1 \\1\end{bmatrix}$, and I got $P = \begin{bmatrix}1 &amp; 1/2 &amp; -2/5\\0 &amp; 1 &amp; 1\\0&amp;0&amp;1\end{bmatrix}$. Then my $P^{-1} = \begin{bmatrix}1 &amp; -1/2 &amp; 9/10\\0 &amp; 1 &amp; -1\\0&amp;0&amp;1\end{bmatrix}$. And I said A's left-multiplication by $P^{-1}$ and right-multiplication by $P (P^{-1}AP)$ gives the diagonal matrix Λ that is similar to A. I found Λ as $\begin{bmatrix}1 &amp; 0 &amp; 0\\0 &amp; 3 &amp; 0\\0&amp;0&amp;21/5\end{bmatrix}$. Then I said that if we multiply both sides of the equation $P^{-1}AP = Λ$ by $P$ from the left and by $P^{-1}$ from the right we get $PP^{-1}APP^{-1} = PΛP^{-1} = A$ and that this is the eigendecomposition of A. Here is my first question: Do you think I understood and did correctly what the asker wanted me to do by saying "Determine the eigendecomposition $A = PΛP^{-1}$."?   I started in the same manner and I got three eigenvectors as  $\begin{bmatrix}x_{11}\\x_{12}\\x_{13}\end{bmatrix}$ = $x_{12} \begin{bmatrix}0\\1 \\0\end{bmatrix}$,  $\begin{bmatrix}x_{21}\\x_{22}\\x_{23}\end{bmatrix}$ = $x_{21} \begin{bmatrix}1\\1 \\1\end{bmatrix}$ and  $\begin{bmatrix}x_{31}\\x_{32}\\x_{33}\end{bmatrix}$ = $x_{33} \begin{bmatrix}1\\-2 \\1\end{bmatrix}$.   I said A is not diagonalizable since the eigenvectors are not linearly independent.  Here is my second question: What the hell does the sentence "Can you write A as a linear combination of rank 1 matrices formed from its eigenvectors?" mean?
98642
I need to verify if the symmetric matrices form a subspace. But I don't know how to represent a general symmetric matrix. I know that the matrix $A$ is symmetric if $A = A^t$ but I can't write a general one and add with another to verify the conditions for a subspace.
62659
I have a conjecture. I have a problem proving or disproving it.      Let $w \in \Lambda^k(V)$ be a $k$-vector. Then $W_w=\{v\in V: v\wedge w = 0 \}$ is a $k$-dimensional vector space if and only if $w$ is decomposable.   For example, for $u=e_1\wedge e_2 + e_3 \wedge e_4$ we have $W_u = 0$.