You probably know both &quot;Householder&quot; and &quot;Eigendecomposition&quot;, but do you really know how it works geometrically? I got stuck in  question asks me to provide the eigen decomposition of householder operator H=I-2ww* where ‖w‖=1, geometrically! I know that the eigen-values are [-1, 1 ...1] but I cannot understand about the eigen-vectors and in particular I cannot depict the solution, geometrically! Can anybody possibly help me to understand it through drawing a geometrical interpretation? Thanks
121659
Define the eigen decomposition of householder operator $H=I-2ww^*$ where $\vert\!\vert w\vert\!\vert=1$, geometrically! I know that the eigen-values are $[-1, 1 \dots 1]$ but I cannot understand about the eigen-vectors and in particular I cannot depict the solution, geometrically! Can anybody possibly help me to understand it through drawing a geometrical interpretation? Thanks alot
78274
The question is if you have a situation where one of the singular values is equal to 0 in a singular value decomposition of a matrix, how to do you procede to find the column vector of U corresponding to this singular value?   Usually I found the columns of U with this relation, so let's say that we refer to $\vec{u_2}$ as the U column vector corresponding to singular value equal to 0:  $\vec{u_2} = (1/s_2)Av_2$  Now the as we can see the 0 is in an illegal place, since you can't divide by zero. But if you move it to the left side, I still can't quite see how to find the column of U corresponding to it. Any suggestions?  Here is an example of a matrix that provided this specific challenge  $ A = \left(\begin{array}{rrr} -3 &amp; 1 \\ 6 &amp; -2 \\ 6 &amp; -2 \end{array}\right). $  $ \lambda{_1} = 90$,  $\lambda{_2} = 0$,  $s_1 = 3 \sqrt{10}$,  $s_2 = 0$,  How can one find $\vec{v_2}$ when $\lambda{_2} = 0$? In this case, you know that the eigenvectors of $A^{t} A$ are orthogonal, solve for $\lambda{_2}$ and find the eigenvector if it's orthogonal to the eigenvector corresponding to the first eigenvalue you have your solution.  So we end up with the following normalized column vectors for V, which we can see are orthogonal:  $ \vec{v_1} = \left(\begin{array}{rrr} -3/\sqrt{10} \\ 1/\sqrt{10} \\ \end{array}\right). $  $ \vec{v_2} = \left(\begin{array}{rrr} 1/\sqrt{10} \\ 3/\sqrt{10} \\ \end{array}\right). $
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?
48103
I am on the proof of the polar decomposition w/o using SVD:  Let $A$ be a real $n \times n$ matrix. Show that there exists $Q$: orthogonal, $P$: positive semi-definite symmetric matrix such that $A = QP$.   Consider the product $A^*A$, it is a self-adjoint and positive semi-definite (as we are in $\mathbb{R}$) symmetric matrix, so it has a decomposition $EDE^{-1}$ where $D$ is diagonal and $E$ is orthonormal and contains eigenvectors of $A^*A$. Now let $$   P = \sqrt{A^*A} =  E\sqrt{D}E^{-1}     $$ where $\sqrt{D}$ has diagonal elements being the squareroot of $D$. Notice that $||Px|| = ||Ax||,  $ so we can expect the product $AP^{-1}$ to be orthogonal. Provided that $P^{-1}$ exists (i.e. $P$ is positive definite), we have $||AP^{-1}x||=||x||$, so we do find our desired decomposition by letting $Q = AP^{-1}$.  The problem is that if $A$ is not invertible, then $P$ will also not invertible, regarding the whole proof useless. How should I fix it in this case?
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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.
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I'm interested in knowing some geometric interpretation of matrices. Can you suggest any lecture note or textbook or anything else about it? I've just finished an undergraduate course in linear algebra, so do not suggest any linear algebra book please, I just want to know more about this topic. Examples of what a mean by "geometric interpretation" are  and .
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I am currently reading two linear algebra books. One is Hoffman/Kunze's and the other one is Friedberg/Insel/Spence's.  They define Jordan canonical form of linear operator in different ways.  In Hoffman's book, Jordan forms are obtained by the primary decomposition and cyclic decomposition. In this case, 1's can appear some sub-diagonal entries. For example,  $M = \begin{array}{cc} 2 &amp; 0 &amp; 0 &amp; 0 &amp; 0 \\ 1 &amp; 2 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 3 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 1 &amp; 3 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 3 \\ \end{array}$  But in Friedberg's (And almost any other linear algebra books for engineers..), 1's can appear some super-diagonal entries. For example,  $M' = \begin{array}{cc} 2 &amp; 1 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 2 &amp; 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 3 &amp; 1 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 3 &amp; 0 \\ 0 &amp; 0 &amp; 0 &amp; 0 &amp; 3 \\ \end{array}$  Can someone explain why they define differently and relation of each other?
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For t∈R, let $A_t = \left( \begin{array}{ccc} t &amp; 1 &amp; 1 \\ 1 &amp; t &amp; 1 \\ 1 &amp; 1 &amp; t \end{array} \right) $. Find the Eigenvalues and Eigenvectors. Is $A_t$ diagonalizable for all t?  So I am thinking of approaching it like this. First, I find the characteristic polynomial of $A_t$ and use that to find the eigenvalues and their corresponding eigenvectors. Then I check if those eigenvectors are linearly independent for all t. If they are linearly independent, then $A_t$ is diagonalizable for all t, and if they're not, $A_t$ is not diagonalizable for all t.  Should I pursue this problem like this? Or is there a more efficient method? Thanks in advance!
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Given a column vector $v$ in $\mathbb{R}^n$, what are the eigenvalues of matrix $vv^T$ and associated eigenvectors?  PS: not homework even though it may look like so.
53254
Let $A$ be a real symmetric matrix. Assume $A^{2}v=\lambda Av$ , can we somehow deduce $v$ is an eigenvector of $A$ with $\lambda$ as its corresponding eigenvalue? i.e $Av=\lambda v$. Thanks Edit: Also assume $\lambda \ne 0$ and we do not know if it is singular or nonsingular.
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I'm learning multivariate analysis and I have learnt linear algebra for two semester when I was a freshman.  Eigenvalue and eigenvector is easy to calculate and the concept is not difficult to understand.I found that there are many application of eigenvalue and eigenvector in multivariate analysis. For example     In principal components, proportion of total population variance due   to kth principal component equal    $$\frac{\lambda_k}{\lambda_1+\lambda_2+...\lambda_k}$$   I think eigenvalue product corresponding eigenvector has same effect as the matrix product eigenvector geometrically.   I think my former understanding may be too naive so that I cannot find the link between eigenvalue and its application in principal components and others.   I know how to induce almost every step form the assumption to the result mathematically. I'd like to know how to intuitively or geometrically understand eigenvalue and eigenvector in the context of multivariate analysis(in linear algebra is also good).  Thank you!
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Question:  Suppose $V$ and $W$ are finite dimensional vector spaces over $\mathbb{C}$. $A$ is a linear transformation on $V$,  $B$ is a linear transformation on $W$. Then there exists a non-zero linear map $C:V\to W$ s.t. $CA=BC$ iff $A$ and $B$ have a same eigenvalue.  ===========This is incorrect==========  Clearly, if $CA=BC$, suppose $Ax=\lambda x$, then $B(Cx)=CAx=C(\lambda x)=\lambda (Cx)$, so $A$ and $B$ have same eigenvalue $\lambda$. On the other hand, if $A$ and $B$ have a same eigenvalue $\lambda$, suppose $Ax=\lambda x, By=\lambda y$. Define $C:V\to W$ s.t. $Cx=y$, then $BCx=By=\lambda y=C\lambda x=CAx$. But I don't kow how to make $BC=CA$ for all $x\in V$.   ======================================
88831
Suppose I am given a $2$x$2$ matrix $A=$ \begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix}  And an initial vector $x_n$ = \begin{pmatrix} x_0 \\ y_0\end{pmatrix}.  Under repeated iteration $x_{n+1} = Ax_n$, we find that $x_{n+1}$ converges to a line in the direction of the $eigenvector$  of $A$ of the form $X$= $[u,mu]$ where $m$ is the slope of the line.  I know that this process somehow also involves the largest $eigenvalue, \lambda,$ of $A$, though I am unsure how it connects.  My question is: How does the largest eigenvalue connect to this repeated iteration, how do I find the slope of the line of convergence $m$, and what is meant by the "direction of the eigenvector"?
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The question is motivated from the following problem:     Let $I\neq A\neq -I$, where $I$ is the identity matrix and $A$ is a real $2\times 2$ matrix. If $A=A^{-1}$, then the trace of $A$ is   $$ (A) 2 \quad(B)1 \quad(C)0 \quad (D)-1 \quad (E)-2$$   Since $A=A^{-1}$, $A^2=I$. If the converse of  is true, then $\lambda^2=1$ and thus $\lambda=\pm1$. And then $\rm{trace}(A)=1+(-1)=0$.   Here are my questions:   Is  $C$ the answer to the quoted problem?  Is the converse of Cayley-Hamilton Theorem, i.e.,"for the square real matrix $A$, if $p(A)=0$, then $p(\lambda)$ is the characteristic polynomial of the matrix $A$" true? If it is not, then what's the right method to solve the problem above?
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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.
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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.
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I've started studying Eigenvector and Eigenvalue.   It says in my book that 0 is excluded from being an eigenvector because it breaks the uniqueness of eigenvalue associated with each eigenvector.   But, there is a proof in my book showing that Eigenspace is a subspace. In order for it to be subspace, does that mean that there must be a zero element? But, Eigenvector can not be zero... Am I misunderstanding something?