What do we call a matrix whose columns are orthogonal, such as $\begin{bmatrix}3 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 2 \\ 0 &amp; 1 &amp; 0\end{bmatrix}$? Is there a special name for it? I tried searching to no avail.
121647
Is there a name for a matrix whose rows (or columns) are non-zero orthogonal vectors ?  It seems to me that "orthogonal matrix" would be a good name, but this is already taken -- it refers to a matrix whose rows (or columns) form an orthonormal set of vectors.
110753
In general inference, why orthogonal parameters are useful, and why is it worth trying to find a new parametrization that makes the parameters orthogonal ?   I have seen some textbook examples, not so many, and would be interested in more concrete examples and/or motivation.
97381
$\omega&gt;0$, Then Compute the matrix:          $e^A$,     where $A= \begin{bmatrix} 0 &amp; \omega \\ -\omega&amp; 0 \end{bmatrix}$      I have never seen any problem like this. So Please help me solving this. Any link for theoretical reading is welcome.
34028
I am reading the book Elements of Statistical Learning and trying to understand singular value decomposition (SVD). In particular, what is an orthogonal matrix as it relates to SVD?  According to Wikipedia, an orthogonal matrix is a square matrix, the transpose of an which is equal to its inverse.   As this CrossValidated question points out however, in the SVD example discussed in Elements of Statistical Learning, this does not hold.   What am I missing?
54795
I'm writing a mathematical library; and I have an idea where I want to automatically turn column matrices and row matrices to vectors, with all of the mathematical properties of a vector.  Answer I'm looking for:  Someone with good mathematical reasoning explaining why:  column matrices, column vectors, row matrices, row vectors should not be treated as the same thing. (The library will ofcourse understand operations like [[1,2],[3,4]] * [1,2], where [1,2] is a vector)  or:  some kind of showcase or example where it is impossible for a library that can't differentiate between row vectors and column vectors to know which one of several possible answers are correct.  or:  some kind of evidence that it is in fact possible to do this.  please note: inner vector multiplication will be easily integrated by using a special function for that function rather than the * sign.
68481
Is it possible to divide a matrix by another? If yes, What will be the result of $\dfrac AB$ if $$         A = \begin{pmatrix}         a &amp; b \\         c &amp; d \\         \end{pmatrix},         B = \begin{pmatrix}         w &amp; x \\         y &amp; z \\         \end{pmatrix}? $$
63231
If $X$ is a singular matrix, then is $X'X$ singular? And why?  (' means transpose)  Thanks in advance!
111898
I'm interested in the properties of zero-diagonal symmetric (or hermitian) matrices, also known as hollow symmetric (or hermitian) matrices.   The only thing I can come up with is that it cannot be positive definite (if it's not the zero matrix): The Cholesky decomposition provides for positive definite matrices $A$ the existence of a lower triangular $L$ with $A=LL^*$. However the diagonal of $LL^*$ is the inner product of each of the rows of $L$ with itself. Since the diagonal of $A$ consists of zeros, so $L$ (and thus $A$) must be the zero matrix.  The sorts of questions that interest me are:   which symmetric matrices can be transformed orthogonally into a zero-diagonal matrix? what can we say about the eigen-values of a zero-diagonal symmetric matrix?. any other interesting known properties??
74623
We have to stochastic variables X and Y, and we define   $ \begin{pmatrix} \tilde{X} \\ \tilde{Y} \end{pmatrix}=\begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix} \begin{pmatrix} X \\ Y \end{pmatrix} $  for $a,b,c,d \in \mathbb{R}$, i.e. $\tilde{X}=aX+bY$ and $\tilde{Y}=cX+dY$. We know that both $X$ and Y, and $\tilde{X}$ and $\tilde{Y}$ are uncorrelated, and the distribution of (X,Y) has density $g(x,y)=f(x)f(y)$. Here $f$ is a density with respect to the lebesgue measure for a probability measure on $\mathbb{R}$, and f is continous and strict positive.      Furthermore, we say that the distribution of (X,Y) is invariant to orthogonal matrices $\begin{pmatrix} a &amp; b \\ c &amp; d \end{pmatrix} $ if $(X,Y)$ and $(\tilde{X},\tilde{Y})$, have equal distributions. In general, a matrix O is orthogonal if $OO^T=O^TO=I$, where I is the identity matrix.   Now I have to show that if the distribution of $(X,Y)$ is invariant to orthogonal transformations, then it holds that  $f(x)f(y)=f(0)f(\sqrt{x^2+y^2})$  for alle $x,y \in \mathbb{R}^2$.   Can anyone help me with this problem?
102335
$$     \begin{bmatrix}     -2 &amp; -2 &amp; 0 &amp; 0 \\     -b &amp; -b &amp; 0 &amp; 0 \\      0 &amp; 0 &amp; -4 &amp; 4 \\      0 &amp; 0 &amp; b &amp; -b \\     \end{bmatrix} $$  I have this matrix and I need to find the values for $b$ that this matrix won't  be diagonalizable.  For what I got this matrix is diagonalizable for every $b$ value.  Please write me also the final answer, thank you very much.
90127
I am interested in describing the group of special orthogonal matrices $SO(n)$ by a set of parameters, in any dimension. I would also like to obtain an expression of the density of the Haar measure in this set of parameters. Could anyone help me on this or indicate a good reference? Thanks
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?
14146
I know that the symmetric matrix has orthogonal eigenvectors corresponding eigenvalues. Also, the eigenvectors from the same eigenvalue are linearly independent.  I need an example of symmetric matrix such that the eigenvectors from the same eigenvalue are not orthogonal.  Also, I know that the eigenvalues of the symmetric matrix are real. But , this statement true iff the entries are real numbers.Right? Since the matrix 2x2 $$A=\begin{bmatrix} 1 &amp;i \\  i &amp;1  \end{bmatrix}$$  has non real eigenvalues
29395
I have a question where I have a matrix $$A=\begin{bmatrix}2&amp;-1\\-1&amp;1\end{bmatrix}$$ and I need to find two examples of a non-zero matrix $X$ such that $AX = XA$. My first matrix is \begin{bmatrix}1&amp;0\\0&amp;1\end{bmatrix} the identity matrix but I cant think of another one. Any ideas?
19400
The question is to find in the space $\mathrm{Mat}_{3\times 2}(\mathbb{\mathbb{R}})$ a non-zero matrix that is orthogonal to $$ A= \begin{pmatrix} 1 &amp; 2\\       3 &amp; 4\\       5 &amp; 6\\ \end{pmatrix}.$$  The inner product is $\langle A, B \rangle = tr(A B^T) = tr(B^T A)$.  Any hints or clues on how to start this?
24837
I am trying to find a basis for the set of all $n \times n$ matrices with trace $0$.  I know that part of that basis will be matrices with $1$ in only one entry and $0$ for all others for entries outside the diagonal, as they are not relevant.  I don't understand though how to generalize for the entries on the diagonal.  Maybe just one matrix with $1$ in the $(1, 1)$ position and a $-1$ in all other $n - 1$ positions?
119950
Problem: Let $A$ and $B$ be real orthogonal matrices, $n$x$n$, where $n$ is an odd number. Prove that, at least one of the matrices $A+B$ and $A-B$ has to be singular.  What have I done so far:  -Since  matrices $A$ and $B$ are real orthogonal, it means that their determinants are $-1$ or $+1$  First I observed matrix $A+B$: $$A+B=AI+BI=ABB^T+BAA^T=AB(B^T+A^T)=AB(A+B)^T\Rightarrow$$  $$det(A+B)=det(AB)det(A+B)^T=det(A)det(B)det(A+B)$$  So, if $detA=detB$ we have $det(A+B)=det(A+B)$ which tells me nothing.  But, if  $detA=-detB$ we have $det(A+B)=-det(A+B)$ which can only hapen if $det(A+B)=0$ making $A+B$ singular.  Then, I tried the same for $A-B$: $$A-B=AI-BI=ABB^T-BAA^T=AB(B^T-A^T)=AB(A-B)^T\Rightarrow$$  $$det(A-B)=det(AB)det(A-B)^T=det(A)det(B)det(A-B)$$  So, if $detA=-detB$ we have $det(A-B)=-det(A-B)$ which can happen if $det(A-B)=0$ making $A-B$ singular.  Is this correct or should I do matrix $A-B$ differently? I am also a little confused why does it have to be said that $n$ is an odd number? What should that tell me?  Any help is greatly appreciated.