Let $D$ be the diagonal matrix and $A^2=D$  So far I've taken 2 approaches, the first being a direct algebraic approach. For a $2 \times 2$ matrix  this reduces to $$  \left( \begin{matrix}     a_{11} &amp; a_{12}  \\     a_{21} &amp; a_{22}  \\     \end{matrix} \right )^2 =  \left( \begin{matrix}     a_{11}^2 +a_{21}a_{12} &amp; a_{12}(a_{11} + a_{22})  \\     a_{21}(a_{11} + a_{22}) &amp; a_{22}^2 +a_{21}a_{12}  \\     \end{matrix} \right ) = \left( \begin{matrix}     d_1 &amp; 0  \\     0 &amp; d_2  \\     \end{matrix} \right )$$ where $d_1 \ne d_2$.   From the top right and bottom left squares it's easy to see that $a_{12}=a_{21}=0$ or $a_{11} +a_{22}=0$   If $a_{12}=a_{21}=0$ then, $$d_1=a_{11}^2 +a_{21}a_{12}=a_{22}^2 +a_{21}a_{12}=d_2$$ which means the eigenvalues aren't distinct. Thus $A$ is diagonal.   This proves it when $A$ is $2 \times 2 $.  For  $n \times n $ $A$, with $n$ linearly independent eigenvectors $\mathbf v_1 , ..., \mathbf v_n$ with corresponding eigenvalues $ \lambda_1 , ... , \lambda_n$ such that for $j \ne k$  $ \lambda^2_j \ne \lambda^2_k$ Now $$ D \mathbf v_k =A^2 \mathbf v_k= \lambda^2_k \mathbf v_k$$   thus $\lambda_k^2$ is and eigenvalue of $D$ and $\mathbf v_k$ is the corresponding basis vector, hence the eigenvectors of $A$ are just the standard basis vectors and hence $A$ is diagonal   Is my conjecture that a square root of a diagonal matrix with distinct eigenvalues is diagonal wrong or is there a general proof?
121133
Assume that there exists an invertible matrix $P$ such that $P^{-1}X^nP$ is a diagonal matrix with distinct eigenvalues, then can I say that $P^{-1}XP$ is also a diagonal matrix with distinct eigenvalues? If so, how do I prove it?
83978
If $A^2 = I$ where $A$ is an $n\times n$ matrix, then why is $\operatorname{rank} (A+I) + \operatorname{rank} (A-I) =n$.   I am not sure why this seems to be true.
9485
Define $T : M_{n×n}(\mathbb{R}) → M_{n×n}(\mathbb{R})$ by $T(A) := A^t$.   I know that the corresponding eigenvalues are $+1$ and $-1$, but I'm not sure how to find the eigenvectors of this transformation, in the case of a $2\times 2$ matrix it's simple, but not in the general case.
18758
Question:     Let $A$ be a $3 \times 3$ Matrix such that $[-3,4,1]$   is the eigenvector corresponding to eigenvalue $3$, and $[6,-3,2]$ is an eigenvector corresponding to the eigenvalue $2$. If $v$ = $[0,5,4]$, compute $A^{2}v$.      Hint: $[6,-3,2]+2[-3,4,1]$ = $v$   Where I'm stuck:  I know that we can  solve A with the equation:   A = $P^{-1}DP$, where $P$ is the matrix consisting of the eigenvectors and $D$ is the matrix consisting of the eigenvalues.  However, since $A$ is a $3$ x $3$, we should have a $3 \times 3$ matrix $D$ and $ 3 \times 3 $ matrix $P$ as well, consisting of the eigenvalues and eigenvectors respectively - and in this case, we only have two eigenvalues and two eigenvectors. Where can I obtain the other one of each?  Any help will be very appreciated.  Thank you.
72135
$A,B,C$ are three $n\times n$ matrices such that $A$ has $n$ distinct eigenvalues, $AB=BA$, and $AC=CA$.   We need to show $BC=CB$.  Let $A$ be diagonalizable with eigenvectors say, $v_1,\dots,v_n$, so there exists $P$ such that $PAP^{-1}=D$. What more I can say from the information? Please help.
2421
As a part of an exercise I have to prove the following:  Let $A$ be an $(n \times m)$ matrix. Let $A^T$ be the transposed matrix of $A$. Then $AA^T$ is an $(n \times n)$ matrix and $A^TA$ is an $(m \times m)$ matrix. $AA^T$ then has a total of $n$ eigenvalues and $A^TA$ has a total of $m$ eigenvalues.  What I need to prove is the following:  $AA^T$ has an eigenvalue $\mu \not = 0$ $\Longleftrightarrow$ $A^TA$ has an eigenvalue $\mu \not = 0$  In other words, they have the same non-zero eigenvalues, and if one has more eigenvalues than the other, then these are all equal to $0$.  How can I prove this?  Thanks and regards.
85458
So the question asks: Let A be a $2\times 2$ matrix, and suppose that there exist scalars $λ_1,λ_2$ in$\mathbb{R}$ and nonzero vectors $v_1,v_2$ in $\mathbb{R}$ such that $Av_1=λ_1v_1, Av_2=λ_2v_2$, and $λ_1$ does not equal to $λ_2$.Prove that the vectors $v_1,v_2$ form a basis of $R^2$.   So so far I got:   As the eigenvector $v_1,v_2$ are corresponding to two different eigen values$λ_1,λ_2$ respectfully.   Therefore the eigenvectors $v_1,v_2$ are linearly independent.   And since $v_1,v_2$ spans A in $\mathbb{R}$.   Hence,  $v_1,v_2$ forms a basis of $\mathbb{R}$.   Does this proof look right? Do I need to write a linear combination of $v_1,v_2$ to show they span A?
76155
If $A\in M_{n}$ has $n$ distinct eigenvalues and if $A$ commutes with a given matrix $B\in M_{n}$, how can I show that $B$ is a polynomial in $A$ of degree at most $n-1$? I think first I need to show that $A$ and $B$ are simultaneously diagonalizable, but I don't know how to begin. Any advice is much appreciated.
86940
Let $X$ be a real $n \times n$ matrix, then there is a Jordan decomposition such that $X = D+N$ where $D$ is diagonalisable and $N$ is nilpotent.  Then, I was wondering whether the following is correct.  $e^{tX}(x) = \sum_{k=0}^{m} \frac{t^k N^k}{k!} \left(e^{t \lambda_1} \alpha_1 v_1+..+e^{t \lambda_n} \alpha_n v_n \right).$  Here $x = \sum_{i=1}^{n } \alpha_i v_i$ and $v_i$ are the eigenvectors of the diagonalisable matrix, $\lambda_i$ are the eigenvalues of $D$ and $m$ is the degree up to which $N^k$ is still non-zero.  Is this correct or am I doing something wrong? Cause I could not find a general equation for this matrix exponential, so I tried my best. (Thus, I am only asking for a verification or correction of this answer.)  If anything is unclear, please let me know.
110840
Let $A$ be an invertible $n\times n$ matrix. If $$Av = \lambda v \qquad (1)$$ for some $v$ and $\lambda$ then $\lambda$ is an eigenvalue of $A$ and $v$ a corresponding eigenvector. Equation $(1)$ may be written as $$Av = \lambda v = \lambda Iv \quad\Leftrightarrow\quad Av - \lambda I v = 0 \quad\Leftrightarrow\quad (A - \lambda I)v = 0$$ where $I$ is the identity matrix. The last equation can be used to find correct $\lambda$'s via the fundamental theorem of algebra (first take determinant).  On the other hand, it is also possible to write $$Av = \lambda v \quad\Leftrightarrow\quad A^{-1}Av = A^{-1}\lambda v \quad\Leftrightarrow\quad Iv - A^{-1}\lambda v = 0 \quad\Leftrightarrow\quad (I - A^{-1}\lambda)v = 0$$ since it was assumed $A$ is invertible.  Now, since $(A - \lambda I)v = 0$ and $(I - A^{-1}\lambda)v = 0$ it must be that $$A - \lambda I = I - A^{-1}\lambda$$ which connects the matrix $A$ to its inverse via the eigenvalues of $A$.  I did some calculations and it seems this works for some invertible matrices $A$ but not all. Am I doing a mistake somewhere?
95101
I hope it is alright to ask something like this here, I am having trouble keeping up with all the special cases and my book is being kind of vague.  I know how to do the standard method of finding diagonal matrices, but I know that triangular matrices are special and my book makes a connection, but it is hard to tell what it is.  Thanks.
16411
Suppose that $A \in \mathbb{R}^{n\times n }$, and $x,y\in \mathbb{R}^n$.    Show that there exist numbers $a,b$ so that $\det{(A + sxy^*)}=a+bs$   Show that if $\det{(A)}\neq 0$ then $a = \det{(A)}$ and $b = \det{(A)} \cdot y^*A^{-1}x$.   Is it true that if $\det{(A)}=0$ then $a=0$?    My answer so far: The third question seems trivial and silly, and I don't understand why it's on there. I have solved the first one as follows: take $\beta$ to be any basis whose first element is $x$. Then with respect to basis $\beta$, $A+ sxy^*$ has the form   $$\left(  \begin{matrix} \alpha_{11} + sy_1 &amp; \alpha_{12} + sy_2  &amp; \dotsb &amp; \alpha_{1n}+sy_n \\ \alpha_{21}&amp;\dotsb&amp;&amp;\alpha_{2n} \\ \vdots &amp; \ddots &amp;&amp;\vdots\\ \alpha_{n1} &amp; \dotsb &amp;&amp;\alpha_{nn} \end{matrix} \right),$$  from which it's clear that the determinant is a polynomial in $s$ of degree $\leq 1$.    I am less clear on the second question.   For a limited case, note that if $A$ is diagonalizable and $x$ is an eigenvector of $A$ with eigenvalue $\lambda_x \neq 0$, then adding $sxy^*$ to $A$ has the effect of preserving all the eigenvectors, bumping up $\lambda_x$ to $\lambda_x + sy^*x$, and since the determinant is the product of all the eigenvalues, the product of $\lambda_1 \lambda_2  \dotsb (\lambda_x + sy^*x)\dotsb \lambda_n $ is   $$\det{(A)} + \det{(A)}\frac{sy^*x}{\lambda_x}=\det{(A)}\cdot sy^*(\frac{x}{\lambda_x}) = \det{(A)}\cdot s y^*(A^{-1}x).$$  Note that $y^*A^{-1}x$ is $\alpha_1 y_1 + \dotsb + \alpha_n y_n$, for $\alpha_i$ such that $x = \alpha_1 A_1 + \dotsb + \alpha_n A_n$, where $A_i$ are the columns of $A$.  But this is all I've got. Can someone give me a hint? I'm not looking for you to feed me the answer. Thanks as always
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.
43790
$\mathbf{A}$ is a $2\times 2$ matrix which satisfies   $\mathbf{A}^2 - 4\mathbf{A} - 7\mathbf{I} = \mathbf{0}$,   where $\mathbf{I}$ is the $2\times 2$ identity matrix. Prove that $\mathbf{A}$ is invertible.      Hint: Find a matrix $\mathbf{B}$ such that $\mathbf{A}\mathbf{B}=\mathbf{I}$.   I tried substituting a variable matrix for $\mathbf{A}$ and substituting $\mathbf{I}$'s value into both the original equation and the hint, but the result was full of equations and didn't seem to help much at all. I would appreciate advice for this problem.  Thanks. --Grace
52902
If $A \in M_{n}(\Bbb{R})$ is an Upper triangular matrix with diagonal entries $1$ such that  $A \neq I$, then what can we say about the diagonalizability of $A$ ?  I know that if the matrix has distinct eigenvalues or the set of eigenvectors are linearly independent then the matrix is diagonalizable.  And that the eigenvalues of the triangular matrices are given by the diagonal elements like  and , but they work nocely if we had distinct elements on the main diagonal.  But in my case I have same value 1 on the main diagonal, how can I approach about the diagonalizability of the matrix?
92627
I have a matrix:  $$ A=\begin{bmatrix}1 &amp; a \\ 0 &amp; 1\end{bmatrix} $$  which satisfies: $A\left( e_{1}\right)=e_{1},A\left( e_{2}\right)=ae_{1}+e_{2}, a\neq 0$  I need compute all powers $A^{n}, n\in \mathbb{Z}$ and find their matrices.  $$ A^n= \begin{bmatrix}1 &amp; na \\ 0 &amp; 1\end{bmatrix} $$  Is it true?  Is there a basis of $V$ (if $V$ is a 2-dimensional real vector space with basis $\left\{e_{1},e_{2}\right\}$) so that the matrix of $A$ is diagonal?
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?