Let S={A: A is an nxn matrix}

Question

Let $S=\{A\in M_{n\times n}: A=A^\tau\}$

$(a)$ Prove S is a Subspace of $M_{n \times n}$, all $n \times n$ matricies

$(b)$ Find a Formula or pattern for dim(S) in terms of n

$(c)$ If A and B are in S, must AB be in S

For part a I just used the 3 step criteria to prove S is a subspace of M

• A =\begin{bmatrix} a & b & c \\ b & e & d \\ c & d & f \end{bmatrix}

• B= \begin{bmatrix} g & h & i \\ h & k & j \\ i & j & l \end{bmatrix}

i) The zero vector in M is in S

ii) For any two vectors in S, the two vectors added are in S

• (A+B)^T=A^T+B^T=A+B

iii) For any scalar r and vector in S, r*vector is in S

• (rA)^T=rA^T=rA

For part B I'm slightly confused. I'm thinking about making a general matrix where everything but the diagonal is 0 and then taking the dimension of that and getting a formula????

• S=\begin{bmatrix} a & 0 & 0 \\ 0 & b & 0 \\ 0 & 0 & c \end{bmatrix}

Then from here just find the dim of S?

For part C I'm using A and B again but changing the matrices

• A =\begin{bmatrix} a & 0 \\ 0 & b \\ \end{bmatrix}

• B =\begin{bmatrix} c & 0 \\ 0 & d \\ \end{bmatrix}

All I did was then do A*B to get

-\begin{bmatrix} a*c & 0 \\ 0 & b*d \\ \end{bmatrix}

and then said "Yes, AB must be in S, since A and B are in S then AB would be a multiple of A and B in S.

(IF IM OVERTHINKING THE MATRICIES IM USING PLEASE UPDATE OR INFORM ME OF AN EASIER MATRIX TO MAKE MY WORK NEATER)

Answer 1

(b) The degrees of freedom are the $n$ on the leading diagonal and the $\binom{n}{2}$ above it, so $\dim S=\frac12n(n+1)$.

(c) $(AB)^T=B^TA^T=BA$, so this fails if $AB\ne BA$. Diagonalizing all but the top $2\times2$ when $n\ge2$, we need only note

$$\left(\begin{array}{cc}1 & 0\\0 & 0\end{array}\right)\left(\begin{array}{cc} 1 & 1\\1 & 0\end{array}\right)=\left(\begin{array}{cc}1 & 1\\0 & 0\end{array}\right).$$