Linear algebra - given a matrix, find a matrix equation that gives information about column orthogonality, their norms and eigenvalues

Question

Given the matrix A below

Find a matrix equation that can give information about:
-orthogonality of column vectors and their norms
-eigenvalues of matrix A including their algebraic multiplicities
-rank of A
-geometric multiplicities of eigenvalues

My try:

If column vectors are orthogonal, then we should get $AA^T = 8I$ (or $A^TA=8I)$, this should tell us that their norms are $\sqrt{8}$.
Multiplying the equation by some eigenvector $x_1$ of $A^T$, we have this:
$A\lambda_1 x_1 = 8Ix_1$
Then this means that for the same (eigen)vector, matrix A has eigenvalue $\frac{8}{\lambda}$
$Ax_1=\frac{8}{\lambda_1}x_1$
What should I obtain from this?Another thing I've noticed is that if we multiply the equation
$A^TA=8I$
by some $x^T$ and $x$ we get
$x^TA^TAx=x^T8Ix$
$(Ax)^TAx=x^T8Ix$
$||Ax||=8||x||$
which tells me that matrix A increases vector norm by 8.

Answer 1

Yes, $A^TA = 8I$ seems pretty good, but you also need $A^T = A$. We get $$\det(A)^2 = \det(A^TA) = \det(8I) = 8^8 \ne 0$$ so $A$ has full rank $8$.

• For canonical vectors $e_i,e_j$ we have $$\langle Ae_i, Ae_j\rangle = \langle A^TAe_i, e_j\rangle = \langle 8e_i, e_j\rangle = 8\delta_{ij}$$ so the columns are pairwise orthogonal and have norm $\sqrt{8}$.
• $A$ is a normal matrix and hence diagonalizable, so the geometric multiplicities are equal to algebraic multiplicities.
• As you noticed, we have $$\|Ax\|^2 = \langle Ax,Ax\rangle = \langle A^TAx, x\rangle = \langle 8x,x\rangle = 8\|x\|^2$$ so if $\|Ax\| = \sqrt{8}\|x\|$. If $Ax = \lambda x$ it follows that $|\lambda| = \sqrt{8}$ so the eigenvalues lie on the circle of radius $\sqrt{8}$ centered at the origin. Also $A$ is symmetric so the eigenvalues are real. Therefore the only possible eigenvalues are $\pm \sqrt{8}$. Both must be present, because otherwise $A$ would diagonalize to $\pm\sqrt{8}I$ which isn't true.
• The equation $\operatorname{Tr} A = 0$ implies that the algebraic multiplicities of $\sqrt{8}$ and $-\sqrt{8}$ are equal (and hence equal to $4$) but you can't get this from $A^TA = 8I$ and $A^T = A$ alone.