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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)

Widawensen
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Displayer124
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  • For ii) see https://math.stackexchange.com/questions/185802/dimensions-of-symmetric-and-skew-symmetric-matrices – Jan Apr 28 '20 at 18:03
  • For (c), here is an idea: https://math.stackexchange.com/questions/3452781/linear-map-matrix-of-transformation/3452790#3452790 – Jean-Claude Arbaut Apr 28 '20 at 18:06

1 Answers1

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(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

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

J.G.
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  • Is that just the format of the answer or is that actually how simple it is? – Displayer124 Apr 28 '20 at 18:57
  • @Displayer124 For (c) you still have to find two symmetric matrices A,B such that AB≠BA. I give a hint above. You have only tested diagonal matrices, but for diagonal matrices AB=BA. – Jean-Claude Arbaut Apr 28 '20 at 19:33