If A is a symmetric matrix, then verify that A×A' (transpose) and also A'×A is also symmetric
121258
I need help with a proof for my liner algebra class.     If $A$ is a square matrix, then $AA^T$ and $A^TA$ are symmetric.   I have no idea where to start!
37668
Let $A$ be an $m\times n$ matrix. Prove that $A^*A + I$ is invertible.   I'm not sure what to do because everything I try ends up at $A^*A$ again and not $AA^*$.
99394
$A$$\rightarrow$$(B$ $\vee$ $C$ ) , $B$ $\rightarrow$ $C$ $\vDash$ $A$ $\rightarrow$ $D$  I think it's wrong but I have no idea how to prove.
36327
How can we prove that the inverse of an upper (lower) triangular matrix is upper (lower) triangular?
108061
Proof: Let eigenvalue $\lambda \neq 0$ such as  $$\textbf{A}\vec{v}  = \lambda\vec{v}$$ $$\Rightarrow (\textbf{A}\vec{v})^\ast = (\lambda\vec{v})^\ast$$ $$\Rightarrow (\vec{v}^\ast\textbf{A}^\ast)=(\lambda^\ast\vec{v}^\ast)$$ Right-multiply both sides by $\color{orangered}{\vec{v}}$$$\Rightarrow (\vec{v}^\ast\textbf{A}^\ast \color{orangered}{\vec{v}} )=(\lambda^\ast\vec{v}^\ast \color{orangered}{\vec{v}} )$$ $$\textbf{A}^\ast=\textbf{A}$$ $$\Rightarrow(\vec{v}^\ast\textbf{A}\vec{v})=(\lambda^\ast\vec{v}^\ast\vec{v})$$ $$\Rightarrow(\vec{v}^\ast\lambda\vec{v}) = (\lambda^\ast\vec{v}^\ast\vec{v})$$ $$\Rightarrow(\lambda\vec{v}^\ast\vec{v}) =  (\lambda^\ast\vec{v}^\ast\vec{v})$$ $$\Rightarrow \lambda = \lambda^\ast$$ $$\Rightarrow \lambda\in\mathbb{R}$$
40159
For any matrix $A$, $AA^T$ is symmetric, and thereby diagonalizable (Spectral Theorem). This implies that $AA^T$ has n linearly independent eigenvectors and therefore full rank.  Does this imply that for any matrix $A$, $AA^T$ is always invertible ? I know there is something wrong here because from the above chain we can conclude that a symmetric matrix is always invertible, which is definitely not true.  Could someone please point out where I am going wrong?
90173
Let $A$ be a $2\times 2$ complex matrix such that $A^2=0$. Prove that either $A=0$ or $A$ is similar over $\mathbb{C}$ to $$\left(\begin{array}{cc} 0 &amp; 0 \\1 &amp; 0 \end{array}\right) $$
22841
Let $A$ and $B$ be two $n \times n$ symmetric and positive definite matrices.   If $A \prec B$, then is it true that $B^{-1} \prec A^{-1}$?  Here, $A \prec B$ means that $B-A$ is positive definite.
60855
For symmetric operators one has: $$A\text{ symmetric}:\quad\mathcal{R}(A\pm\imath)=\mathcal{H}\implies A^*=A$$ How to prove this in an unveiling way?
18721
Let A,B,C,D be complex matrices $n \times n$ such that $AB^T,CD^T$ are symmetric and $AD^T-BC^T=I$. Prove that $A^TD-C^TB=I$. Can anyone give me any idea? Thank you.
7978
Is there any proof of the fact that the eigenvalues of symmetric matrices (i.e. $A\in\mathbb{R}^{n\times n}$ with $A^t=A$) are real without the use of the concept of complex numbers?
99426
Let $A=$ $\begin{pmatrix} 2 &amp; 1 &amp; -1\\ 3 &amp; 0 &amp; 1\\ -2 &amp; 5 &amp; 3 \end{pmatrix}.$  Find a symmetric matrix $B$ such that for every $3\times1$ matrix $\text{x}$,   $$\text{x}^TA\text{x}=\text{x}^TB\text{x}$$  I've tried to take the inverse of $A$ and $\text{x}$ to cancel out, but I get nowhere.   Thanks.
36172
Let $A$ be a $m\times n$ real matrix. If for two suitable matrices $B$ and $C$, $$AA^TB=AA^TC$$ can we conclude that $A^TB=A^TC$?  I know $AA^T$ is a symmetric matrix. Every real symmetric matrix is diagonalizable. But I am not sure how to use this information. Thanks for reading and helping me out with hints.
116982
If $(I-A)(I+A)^{-1}$ is orthogonal then prove that A is skew symmetric.  I can't solve this question.
16624
A matrix is diagonalizable iff it has a basis of eigenvectors. Now, why is this satisfied in case of a real symmetric matrix ?
47179
Let K be nonsingular symmetric matrix, prove that if K is a positive   definite so is $K^{-1}$ .   My attempt:  I have that $K = K^T$ so $x^TKx = x^TK^Tx = (xK)^Tx = (xIK)^Tx$ and then I don't know what to do next.
48254
Are there matrices such that $AB=I$ and $BA \neq I$ ?   $A$ and $B$ are square matrices