Let $A$ be an orthogonal matrix, i.e. $A^T=A^{-1}$. Prove that $A(x\times{y})=\det(A)(Ax\times{Ay})$. I know that orthogonal matrices preserve distance, angles and orthogonality of vectors and I have tried using this idea in conjunction with the formula $|x\times{y}|=|x||y|sin\theta(x,y)$.  (Side-question: How do we know orthogonal matrices preserve distances, angles and orthogonalit of vectors?)
20938
How can one prove the following identity of the cross product?  $$(M a)\times (M b)=\det(M) (M^{\rm T})^{-1}(a\times b)$$  $a$ and $b$ are 3-vectors, and $M$ is an invertible real $3 \times 3$ matrix.
96994
Let $A \in M_{n×n}(\mathbb{R})$ be a skew-symmetric matrix, i.e., $A^t = −A$. Prove that if $n$ is odd, then $\det{A} = 0$.
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?
60530
I would like a proof of the following result, given on .  For all square matrices $\mathbf{A}$ and column vectors $\mathbf{u}$ and $\mathbf{v}$ over some field $\mathbb{F}$, $$ \det(\mathbf{A}+\mathbf{uv}^\mathrm{T}) = \det(\mathbf{A}) + \mathbf{v}^\mathrm{T}\mathrm{adj}(\mathbf{A})\mathbf{u}, $$ where $\mathrm{adj}(\mathbf{A})$ is the adjugate matrix of $\mathbf{A}$.  Note that $\mathbf{A}$ may be singular. However, the proof given on wikipedia requires that $\mathbf{A}$ is nonsingular.
41568
Let $a$ and $b$ be two unitary vectors in $\mathbb E^3$, and let $Q$ be an orthogonal matrix. Does the following hold?  $$Qa \wedge Qb = \pm Q(a \wedge b)$$
107241
Let $A$ and $B$ are $n\times n$ matrices and $x \in C^{n}$.  If $Ax = Bx$ for all $x$ then $A = B$. To prove this I have selected  $x$ from Euclidean basis B = {$e_{1},e_{2},...,e_{n}$}. Then $Ae_{i} = Be_{i}$ implies $i^{th}$ column of A = $i^{th}$ column of B for all $1 \leq i \leq n$. Hence $A = B.$    Is this proof complete?
15798
If $A$ is a $n\times m$ matrix, is the formula $(\text{ker }A)^\perp=\text{im }A^T$ necessarily true?  I'm thinking that rank-nullity would be the simplest and easiest way to prove this, but would applying the definition of linear independence to the components of $(A^T)$ be more illustrative? I'm a bit hung up on this "necessity" issue, and not sure if it has a particular logical meaning in linear algebra.
40247
So far, I've deduced that if the rank of A is n, then all the columns of A are linearly independent since A has n columns. As a result, m must be greater than or equal to n.  In the case that m = n, then A is a square matrix and we can just use matrix operations to distribute the inverse operation over the parenthesis; then we can apply the associative property and compute A*A-1 and AT-1 * AT first, eventually giving that the expression inside the norm is just the identity matrix, where the 2-norm of the identity matrix is obviously 1.  I'm having trouble for the case where m > n. I've identified that the expression inside the norm is actually the matrix of orthogonal projection, but I'm not sure if that will help me with the proof. Help is greatly appreciated, thanks.  This problem is from the text of "Matrix Computations" 3rd edition by Golub after covering a section on orthogonality and SVDs for some background.
83594
Let $A$ be an $n\times n$ matrix. Then its characteristic polynomial is  $$\phi_A(x) = \det(xI - A) = x^n - (\operatorname{tr} A)x^{n-1} + \cdots + (-1)^{n-1}(\operatorname{tr}(\operatorname{adj}A))x + (-1)^n\det A,$$  where $\operatorname{adj}A$ denotes the  of $A$.     My question: Why are the coefficients of $\phi_A$ those shown above?   I can see why $(-1)^n\det A$ is the constant term, since $$\det(A) = \phi_A(0) = (0-\mu_1)\cdots(0-\mu_n) = (-1)^n\mu_1\cdots\mu_n,$$  where $\mu_i$ are the roots ($i=1,\dots,n$).  I also see why the coefficient of $x^{n-1}$ is minus the trace, since by Laplace expansion (along the first row):  \begin{align*}     &amp;\det(xI-A)\\[10pt]    &amp;= \begin{vmatrix}          x-a_{11} &amp; -a_{12} &amp; \cdots &amp; -a_{1n}\\          -a_{21} &amp; x-a_{22} &amp; \cdots &amp; -a_{2n}\\          \vdots &amp; \vdots &amp; \ddots &amp; \vdots\\          -a_{n1} &amp; -a_{n2} &amp; \cdots &amp; x-a_{nn}     \end{vmatrix}\\[10pt]     &amp;= (x-a_{11})\begin{vmatrix}          x-a_{22} &amp; \cdots &amp; -a_{2n}\\          \vdots &amp; \ddots &amp; \vdots\\          -a_{n2} &amp; \cdots &amp; x-a_{nn}     \end{vmatrix} + \underbrace{a_{12} \begin{vmatrix}          -a_{21} &amp; \cdots &amp; -a_{2n}\\          \vdots &amp; \ddots &amp; \vdots\\          -a_{n1} &amp; \cdots &amp; x-a_{nn}     \end{vmatrix} + \cdots}_{\text{at most polynomial of degree $n-2$}}\\[10pt]    &amp;=(x-a_{11})\bigg((x-a_{22})\begin{vmatrix}         x-a_{33} &amp; \cdots &amp; -a_{3n}\\        \vdots &amp; \ddots &amp; \vdots\\        -a_{n3} &amp; \cdots &amp; x-a_{nn}     \end{vmatrix}     +\underbrace{\underbrace{a_{23}\begin{vmatrix}         -a_{32} &amp; \cdots &amp; -a_{3n}\\        \vdots &amp; \ddots &amp; \vdots\\        -a_{n2} &amp; \cdots &amp; x-a_{nn}     \end{vmatrix}+\cdots}_{\text{at most polynomial of degree $n-3$}}      \bigg) + \cdots}_{\text{at most polynomial of degree $n-2$}}\\     &amp;\kern10pt\vdots\\[10pt]     &amp;=(x-a_{11})(x-a_{22})\cdots(x-a_{nn}) + \underbrace{\cdots}_{\text{at most polynomial of degree $n-2$}}\\[10pt]     &amp;= x^n - (a_{11}+a_{22}+\cdots + a_{nn})x^{n-1} + \cdots\\[10pt]     &amp;= x^n - (\operatorname{tr}A)x^{n-1} + \cdots \end{align*} (Let me know if there is a more elegant way to obtain this).   But what I cannot seem to prove is that the coefficient of $x$ is $ (-1)^{n-1}\operatorname{tr}(\operatorname{adj}A)$. Attempting to work with the Laplace expansion (as I did to obtain the trace above) is leading me nowhere.
21095
I was hoping that someone could point me in the right direction on how I could start answering the following question:     Based on the fact that $\det(AB)=\det(A)\det(B)$ for all square $n \times n$ matrices $A,B$, explain in every detail why:   $$\det(A^{-1}) = 1/\det(A).$$
77656
Let $S$ be the subspace , of  $M_{n \times n}(\mathbb R)$ (the vector space of all $n \times n$ real matrices ) , generated by matrices of the form $AB-BA$ , where $A,B \in M_{n \times n}(\mathbb R)$ , then how do we prove that $\dim S=n^2-1$ ? The only thing that I can determine is that the trace of all matrices of $S$ is $0$ . Please help
78573
Well considering two $n \times n$ matrices does the following hold true: $$\det(A+B) = \det(A) + \det(B)$$ Can there be said anything about $\det(A+B)$?  If $A/B$ are symmetric (or maybe even of the form $\lambda I$) - can then things be said?
11744
$A$ is a real orthogonal matrix and $(I + A)$ is non-singular. Prove that $ (I+A)^{-1}(I-A)$ is a skew-symmetric matrix.  Attempt:  $[(I+A)^{-1}(I-A)]^t=(I-A)^t((I+A)^{-1})^t=(I-A)^t((I+A)^{t})^{-1}=(I-A^t)   ((I+A^t))^{-1}$   Please help me further.
85203
Let $A, B$ be $n\times n$ with $n\ge 2$ nonsingular matrices with real entries such that $$A^{-1} + B^{-1} =(A+B)^{-1}$$   then prove that $\operatorname{det}(A)=\operatorname{det}(B)$.   Also show that this result is not valid for complex matrices.   I'm not getting any way out to solve this. Can anybody solve the problem? Thanks for assistance in advance.
68836
I was trying to show that orthogonal matrices have eigenvalues $1$ or $-1$.  Let $u$ be an eigenvector of $A$ (orthogonal) corresponding to eigenvalue $\lambda$. Since orthogonal matrices preserve length, $ \|Au\|=|\lambda|\cdot\|u\|=\|u\|$. Since $\|u\|\ne0$, $|\lambda|=1$.  Now I am stuck to show that lambda is only a real number. Can any one help with this?
7461
Let $A$ be an $n \times n$ matrix and $k$ be a scalar. Prove that $\det(kA)=k^n\det A$.   I really don't know where to start. Can someone give me a hint for this proof?
83671
Let $A$ be $m \times n$ matrix, and $B$ be $n \times m$ matrix. Then  Show that $$\det\begin{bmatrix}I_{n} &amp; B\\ A &amp; I_{m} \end{bmatrix}=\det\begin{bmatrix}I_{m} &amp; A\\ B &amp; I_{n}\end{bmatrix}$$  Show that $$\det(I_{m}-AB)=\det(I_{n}-BA)$$  Is it true that $I_{m}-AB$ and $I_{n}-BA$ have the same rank?    Someone please show me a way to start proving. I tried to use formulas for calculating block matrix with square matrices, but it didn't help me on this one. I really don't know where to start.