Assume that $A$ is an $n\times n$ matrix. Assume that $u$ and $v$ are eigenvectors corresponding to eigenvalues $5$ and $9$. Determine all pairs of real numbers $α$ and $β$ such that $αu+βv$ is also an eigenvector of $A$.  Could someone please advise if this works! Since u and v are eigenvectors corresponding to eigenvalues 5 and 9,   We have Au = 5u …(1)       Av = 9v  …(2) Now we want to find all pairs of α and β  lets call them  α_i  and β_i  such that (α_i u+β_i v)  is eigen vector of A i.e.           $$      A(α_i u+β_i v)=e_i (α_i u+β_i v)    &lt; => Aα_{i} u+Aβ_{i} v=e_{i} (α_{i} u+β_{i} v)    &lt; => α_{i} Au+β_{i} Av=e_{i} (α_{i} u+β_{i} v)     &lt; => α_{i} 5u+β_{i} 9v=e_{i} (α_{i} u+β_{i} v)     &lt; => e_{i}=((α_{i} 5u+β_{i} 9v))/((α_{i} u+β_{i} v) )   …(3)   e_{i} $$ will be=k when         $$ α_{i}$$  happens to be the multiple of 9 i.e.$$α_{i}=9*k,        β_{i} $$ happens to be multiple of 5 i.e.5*k   except when k = 0   Thus for$$ (α_{i} u+β_{i} v) $$ to be eigenvector of A $$ (α_{i},β_{i} )=[(9*k,5*k)]  ∀ k = ±1,±2,±3$$…and k!=0    So my question is different from the possible duplicate highlighted below - as it has a specific linear combination to deal with
121835
I believe I am very close to finishing this proof, but I cannot figure out the last part. If anybody could check my work and maybe give me a little hint, it would be greatly appreciated!  Let $V$ be a finite-dimensional vector space and $T \in \mathcal{L}(V)$, and let $\mathbf{u,v} \in V$ be eigenvectors of $T$.  Claim:  If $\mathbf{u} + \mathbf{v}$ is an eigenvector of $T$, then $\mathbf{u}, \mathbf{v}$, and $\mathbf{u+v}$ all correspond to the same eigenvalue.  Proof (So far!): Suppose $T(\mathbf{u}) = \lambda_1 \mathbf{u}$ and $T(\mathbf{v}) = \lambda_2 \mathbf{v}$ with $\mathbf{u}, \mathbf{v} \neq \mathbf{0}$.   Now suppose $T(\mathbf{u+v}) = \lambda_3(\mathbf{u+v})$ with $\mathbf{u+v}\neq \mathbf{0}$.  Then, $$ T(\mathbf{u}) + T(\mathbf{v}) = \lambda_3 \mathbf{u} + \lambda_3 \mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} =  \lambda_3 \mathbf{u} + \lambda_3\mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} - \lambda_3 \mathbf{u} -\lambda_3\mathbf{v} = \mathbf{0}\\ (\lambda_1 - \lambda_3) \mathbf{u} + (\lambda_2 - \lambda_3)\mathbf{v} = \mathbf{0} $$  Now I know in order to show that $\lambda_1 = \lambda_2 = \lambda_3$, I must show that the only solution to the last line is the trivial one. This would imply that $\mathbf{u}$ and $\mathbf{v}$ are linearly independent which I am unconvinced of!  The only information I have to my advantage I haven't used yet is the fact that $\mathbf{u}, \mathbf{v}, \mathbf{u+v} \neq \mathbf{0}$. I really cannot see how this information can help me though.  Perhaps I am going about this wrong, but that's why I want to ask! Thanks for your help.
85937
Prove:     If $v_1,...,v_r$ are eigenvectors that correspond to distinct eigenvalues $\lambda_1, ...,\lambda_r$ of an $n \times n$ matrix $A$, then the set $\{v_1,...,v_r\}$ is linearly independent.   Please give an example and tell me how this theorem works!
91498
So I was working on a specific problem related to Hermitian matrices. If we let $H_n$ denote the set of n x n Hermitian matrices. We're told that $H_n$ is a real vector space under matrix addition and scalar multiplication by a real number. I don't understand why though, because what would prevent you from adding two imaginary numbers?   Then we're told to write a basis for $H_2$. Now for a Hermitian matrix, it's equal to its conjugate transpose, so a basis would be:  $$\left(\begin{array}{ccc} 0 &amp; 1 \\ 1 &amp; 0\\ \end{array}\right) , \left(\begin{array}{ccc} 0 &amp; 0 \\ 0 &amp; 1\\ \end{array}\right), \left(\begin{array}{ccc} 1 &amp; 0 \\ 0 &amp; 0\\ \end{array}\right), \left(\begin{array}{ccc} 0 &amp; i \\ -i &amp; 0\\ \end{array}\right)$$  Which should be the same as the basis for a symmetric matrix, correct?   From there though, we have $sH_n$ represent skew-Hermitian matrices. I believe the basis for $sH_n$ is:  Edit: updated my basis here.   $$\left(\begin{array}{ccc} 0 &amp; -1 \\ 1 &amp; 0\\ \end{array}\right), \left(\begin{array}{ccc} 0 &amp; i \\ i &amp; 0\\ \end{array}\right)$$  because the diagonals would have to be zero right, to make the conjugate transpose of A be equal to -A?  From there though, they ask if $sH_n$ is a real vector space, and I'm not sure what to answer. It's real if we only use reals?  I'm asked the same thing for $U_n$, where $U_n$ represents n x n unitary matrices. I'm not even sure how to come up with a basis for this set??  Edit: Technically, the question says "Is $U_n$ a real vector space? Write a basis if possible."  I suppose I couldn't write a basis because $U_n$ is not a real vector space then. How would I show that though?   Thanks in advance for your help.
14097
I am given that $V$ is n-dimensional vector space over $\mathbb{C}$ and $T \in L(V)$. And $T$ has least $m$ distinct nonzero eigenvalues. How do I show that $\text{null}(T^{n-m}) = \text{null}(T^{n-m+1})$?  I'm really not sure how to start here so just the first couple steps would be very helpful. Thanks!!
82698
The proof given by my textbook is highly non-satisfying. The author adopted some magic-like "reductio ad absurdum" and the proof (although is correct) didn't reveal the nature of this problem. I made my own effort into it and tried a different approach. Yet I can't finish it.  Let $\mathscr{F}$ be a commuting family in $M_n(\mathbb{C}^n)$, and $A\in\mathscr{F}$, then $A$ has $n$ eigenvalues. We pick one, say $\lambda$. Let $x$ be one of its eigenvector.  We can easily prove that, if $A$ has no other eigenvector with eigenvalue $\lambda$ that linearly independent with $x$, which means that $\{cx|c\in\mathbb{C}\}$ are the only vectors satisfying $Ax=\lambda x$, then $x$ is a common eigenvector. Because $\forall B\in\mathscr{F}$ and $\forall y\in \{cx|c\in\mathbb{C}\}$, $$A(Bx)=ABx=BAx=B(Ax)=B(\lambda x)=\lambda (Bx)$$, so that $Bx$ has to be in $\{cx|c\in\mathbb{C}\}$, that is, $Bx=c_0x$ for some $c_0\in\mathbb{C}$, which means $x$ is a eigenvector of $B$ too.  But what if there are vectors satisfying $Ax=\lambda x$ that's not in  $\{cx|c\in\mathbb{C}\}$? Well, then we should have a set of linearly independent eigenvectors $\{x_1,x_2,...,x_k\}$, that $\{c_1x_1+c_2x_2+...+c_kx_k|c_i\in\mathbb{C}\}$ are the only vectors satisfying $Ax=\lambda x$.  Now, I have a reasonable hypothesis that there exists some $x=c_1x_1+c_2x_2+...+c_kx_k$, that can be proven to be a common eigenvector of $\mathscr{F}$. I've tried some approaches to prove it but all failed.  Do you guys believe it's true? And if it is true then how do I prove it?
63156
Let $T$ be a self adjoint operator on a finite dimensional inner product space. And let $4$ and $5$ be the only eigenvalues of $T$. Show $T^2-9T + 20I = 0$ .  proof:Let $T$ be a self adjoint. And let $4 , 5$ be the only eigenvalues of $T$. Then $Tv = 4v, Tv = 5v$ for some $v \neq 0$. Then $Tv = 4v, Tv = 5v$ if and only if $(T-4I)v = 0, (T-5I)v=0$.  Then $ (T-4I) (T-5I)v=0$ implies $(T^2 -5T -4T + 20I^2)v = 0$, implies   $(T^2 -9T + 20I)v = 0$, so we have  $T^2 -9T + 20I = 0  $. Can someone please check this? if it's not correct, could I please have some help on it? Thank you
8575
$A$ is an $n\times n$ matrix of complex numbers. Prove that if $\lambda$ is an eigenvalue of $A^2,$ then $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of $A.$    If $\lambda$ is an eigenvalue of $A^2,$ we have $\lambda\alpha=A^2\alpha$ for some $\alpha.$ Then how can we find a $\beta$ s.t. $\sqrt{\lambda}\beta=A\beta?$
52205
This question is in reference to  proof of the Spectral Theorem.   Let $A$ be a symmetric matrix. Take $u$ to be an orthonormal vector (such that $u^Tu=1$). Then if $u$ is an eigenvector of $A$, then $Au=\lambda u$. This implies that $\lambda=u^TAu$.   The article says that this proves that $\lambda$ is real. I don't follow why this is. Both $u$ and $A$ can contain complex entries! Why would $u^TAu$ have to be real?
58133
Firstly, I know this question might seem slightly vague, but I can't find any examples or specific questions to specify it.   My question is in regards to finding the Jordan Canonical Form of a matrix, more specifically when you have two Jordan chain of length greater than $1$ for a particular eigenvalue. After you've found the eigenvectors, are these eigenvectors always linearly independent? Also, if they are, is it possible to find eigenvectors which form Jordan chains which are 'wrong' because they aren't linear independent?
48334
So $Ax=\lambda x$ where $A$ is the matrix, $\lambda$ and $x$ its eigenvalue and eigenvector resptectively.  Now I have another matrix similar to $A$ i.e. $B=TAT^{-1}$  I'll let a vector $y$ which is an eigenvector of $B$, i.e. $By=\mu y$ where $\mu$ is its ($B$'s) eigenvalue.  I have to now prove that  $By=\mu x$.  Some maths leads me to  $By = TAT^{-1}y$ but here I'm stuck.  answers wants me to assume $y=Tx$ which will essentially solve it but why/how is $y=Tx$?
101802
Show that if $A,B \in M_{n \times  n}(K)$, where $K=\mathbb{R}, \mathbb{C}$, then the matrices $AB$ and $BA$ have same eigenvalues.   I do that like this:  let $\lambda$ be the eigenvalue of $B$ and $v\neq 0$  $ABv=A\lambda v=\lambda Av=BAv$  the third equation is valid, because $Av$ is the eigenvector of $B$. Am I doing it right?
78303
$$u = (r,1,1), \ v = (1,s,2s), \ w = (1,2,2)$$  I know to assume the set will be linearly dependent, equate them to the $0$ vector and find the values not equal to what $r$ and $s$ is. But I can't seem to reduce the matrix to echelon form.  I've gotten to the point of:  $$     \begin{pmatrix}     r &amp; 1 &amp; 1 \\     1 &amp; s &amp; 2 \\     1 &amp; 2s &amp; 2 \\     \end{pmatrix} $$  $R_2 = R_2 - R_3$ $$     \begin{pmatrix}     r &amp; 1 &amp; 1 \\     1 &amp; -s &amp; 0 \\     1 &amp; 2s &amp; 2 \\     \end{pmatrix} $$ $R_3 = R_3 +2R_2$  $$     \begin{pmatrix}     r &amp; 1&amp; 1 \\     0 &amp; -s &amp; 0 \\     1 &amp; 0 &amp; 2 \\     \end{pmatrix} $$  EDIT: Thank you @LordSharktheUnknown  $R_3 = r \times R_3 - R_1$  $$     \begin{pmatrix}     r &amp; 1 &amp; 1 \\     0 &amp; -s &amp; 0 \\     0 &amp; -1 &amp; 2r-1 \\     \end{pmatrix} $$  And I'm stuck. I know $s ≠ 0$ for it to be linearly independent but I have no idea how to reduce the rows any further to get to $r$ by itself.  Any help is appreciated.  Thank you.
39494
Struggling to find a place to start with this proof- just began learning about Jordan normal.  Given a 2-by-2 matrix $A$ and a Jordan normal form matrix $J_{\lambda}$, there exists a matrix $S = [v1, v2]$ s.t. $$S^{-1}AS = J_{\lambda}$$ That is, $J_{\lambda}$ is similar to $A$.  WTS: $$(A - \lambda I)v_{1} = 0 $$ $$(A - \lambda I)v_{2} = v_{1}$$    I was going to try to start with the similarity condition and work into producing the two statements below.  Transform the similarity condition to: $$(A = SJ_{\lambda}S^{-1}) * v_{1}$$  Rewrite our first statement to prove as: $$Av_{1} = \lambda I v_{1}$$   Substitute: $$\lambda I v_{1} = SJ_{\lambda}S^{-1}v_{1}$$  But I've been continually getting stuck.  Tried a few ways other than this.  What intuition am I missing?
45709
We were asked to prove the following:  Let $ A $ be an $n \times n$ matrix with $n$ distinct real eigenvalues. If $AB=BA$, show that $B$ is diagonalizable.   It was suggested I show that an eigenvector of $A$ is also an eigenvector of $B$. I am both having trouble doing this and failing to see how I would complete the proof after. Any help would be appreciated.  Thanks
52271
I have stumbled across a very old exam question from my linear algebra course and the solutions are not available. I was wondering if my working/logic is correct and if any improvements can be made.     Let  $\ B=\begin{pmatrix}     1 &amp; -1 &amp; 4 \\     6 &amp; -7 &amp; 2 \\     -3 &amp; 1 &amp; -6       \end{pmatrix}\ $ and $\ \vec{v_1}=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.      Part (i): Calculate $(B+5I)\vec{v_1}$ and $(B+5I)^2\vec{v_1}$.   Using simply matrix multiplication, I calculated $(B+5I)\vec{v_1}=\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and $(B+5I)^2\vec{v_1}=\vec{0}$.     Part (ii): Find all eigenvalues and eigenvectors of B.   From our calculations in part (i), we can see that $\ \lambda=-5,-5$ are two of the eigenvalues, with corresponding eigenvector $\ \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}$ and generalised eigenvector $\ \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. From the trace of B, the remaining eigenvalue is $\lambda=-2$ with corresponding eigenvector $\begin{pmatrix} -2 \\ -2 \\ 1 \end{pmatrix}$.     Part (iii): Giving reasons, write down a basis for the generalised eigenspace $GE_{-5}$ of B.   I am unsure of this part of the question. If I had to guess, I would say a basis for $GE_{-5}=\left\{\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix},\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}\right\}$. Is this correct?     Part (iv): Solve the initial value problem $\vec{y'}=B\vec{y}$ where $\ B=\begin{pmatrix}     1 &amp; -1 &amp; 4 \\     6 &amp; -7 &amp; 2 \\     -3 &amp; 1 &amp; -6       \end{pmatrix}\ $ and $\vec{y}(0)=\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$.   Well, the solution to this IVP will take the form $\vec{y}=e^{tB}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$. I noticed that $\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}$ was a generalised eigenvector and hence we can write $\vec{y}=e^{-5t}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}+e^{-5t}\begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix}t$. Thus $\vec{y}=e^{-5t}\begin{pmatrix} 1+t \\ 1+2t \\ -1-t \end{pmatrix}$
24766
Let $A\neq I$ be a $5\times5$ matrix with real entries such that the sum of the entries in each row of $A$ is $1$. Then the sum of all the entries in $A^3$ is  1)$\space 3$ $\qquad $2)$\space 15$ $\qquad$ 3)$\space 5$ $\qquad$ 4)$\space 125$  Solution:  The answer is $5$ if $A = I$, but for $A \neq I$,  I found that vectors $\begin{pmatrix} 1\\ 1\\ 1\\ 1\\ 1\\ \end{pmatrix}$ and  $\begin{pmatrix} -1\\ -1\\ -1\\ -1\\ -1\\ \end{pmatrix}$  are eigen vectors for the matrix $A$ corresponding to eigen value $1$. From here, I dont know how to proceed, please help me solve the question..and if there are mistakes in my argument, please tell me..
8100
If I have a $2\times 2$ or $3\times 3$ matrix, how should I go about computing the eigenvalues and eigenvectors of the matrix?  NB: I am making this question to provide a unified answer to questions about eigenvalues of small matrices so that all of the specific examples that come up can be marked as duplicates of this post. See .