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Question

Consider the linear operator $P$ defined on $M_{n}(\mathbb{R})$ by

$P(A)$ = $\frac{1}{2}(A+A^{T})$

Show that $P$ is a projection, and describe its image and kernel.

I have shown that $P$ is a projection, but I'm a bit unsure about what is meant by "describe its image and kernel". The image and kernel of P are given by the sets:

$Im(P)$ = {$P(A)$ $∣$ A $\in$ $M_{n}(\mathbb{R})$}

$Ker(P)$ = {$A$ $\in$ $M_{n}(\mathbb{R})$ $∣$ $P(A)$ = $0$}

Is there some sort of relationship between these sets that I'm supposed to describe?

Thanks in advance.

Martin Sleziak
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Joey
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1 Answers1

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Projection:$P^2(A) = P(P(A)) = P(\frac{1}{2}(A + A^T)) = \frac{1}{2}(\frac{1}{2}(A + A^T) + \frac{1}{2}(A + A^T)^T) = \frac{1}{4}(A + A^T + A^T + A) = \frac{1}{2}(A + A^T) = P(A)$

Kernel: $P(A) = 0 \implies \frac{1}{2}(A + A^T) = 0 \implies A = -A^T$ i.e. anti-symmetric matrices

Kernel($P$) = {$A\in M_n(\mathbb{R}) : A=-A^T$}

Image: $\frac{1}{2}(A + A^T)$ is a symmetric matrix (since $(\frac{1}{2}(A + A^T))^T = \frac{1}{2}(A^T + A)$)

Image($P$) = {$A\in M_n(\mathbb{R}) : A=A^T$}

  • How do you arrive at $A$ = $A^{T}$ as a condition for the set of images of $P$? – Joey Sep 24 '15 at 16:25
  • $B = \frac{1}{2}(A + A^T)$. See that $B = B^T$ –  Sep 24 '15 at 18:09
  • But how does that imply that $A$ = $A^{T}$? Shouldn't the Image of $P$ be instead: Image($P$) = {$B$ $\in$ $M_{n}(\mathbb{R})$ : $B$ = $B^{T}$ }, where $B$ = $\frac{1}{2}(A+A^{T})$ ? – Joey Sep 24 '15 at 20:24
  • We can write any symmetric matrix in the form of $\frac{1}{2} (A + A^T)$. Let S is a symmetric matrix, then $S =\frac{1}{2} (S + S) = \frac{1}{2} (S + S^T)$ since $S = S^T$. This implies that the image is the set of all symmetric matrices –  Sep 24 '15 at 20:35
  • That makes more sense. Thank you so much for your help! – Joey Sep 24 '15 at 20:47