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$T:\mathbb{R}_{2\times 2} \to \mathbb{R}_{2\times 2}$ such that $T(A)=\frac{1}{2}(A+A^t),~~ A\in \mathbb{R}_{2\times 2}$. show that T is linear, find $Ker~ T$, $Im ~T$

Attempt:

$T(A+B)=\frac{1}{2}(A+B+(A+B)^t)=\frac{1}{2}(A+A^t+B+B^t)=T(A)+T(B)$ Similarly $T(cA)=cT(A)$. So T is linear.

I can find $Ker~ T=\{A\in \mathbb{R}_{2\times 2}/T(A)=O\}=\{A\in \mathbb{R}_{2\times 2}/A=-A^t\}$ but unable to find $Im ~T$.

How to find $Im ~T$?

Martin Sleziak
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user1942348
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2 Answers2

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$\operatorname{Ker}T$ is all Skew Symmetric Matrices..and $\operatorname{Im}T$ is all symmetric Matrices...Because every symmetric Matrix can be expressed as $1/2(A+A')$ .

For any Symmetric Matrix $A = A'$ so any symmetric Matrix $A = 1/2(A+A')$. So every Symmetric Matrix of order $2$ has an preimage . So set of all symmetric Matrices of order two $S$ is a subset of $\operatorname{Im}T$.

Now we have to show $\operatorname{Im}T$ is a subset of $S$. Now we will check whether there exists any $A$ for which $T(A)$ is not Symmetric. Every $T(A)$ is $1/2(A+A')$ which is a Symmetric Matric. So Set of all Symmetric Matrices of order $2$ is $\operatorname{Im}(T)$.

Martin Sleziak
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anonymous
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  • I know that $Im(T)={T(A)/A\in \mathbb{R}{2\times 2}}={1/2(A+A')/A\in \mathbb{R}{2\times 2} }$. Please elaborate, how you have claimed? – user1942348 Apr 04 '17 at 10:35
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    Hiii@user1942348 This is the definition of $T(A)$ u made in ur qus...Btw sorry for late reply...I did not get notified... – anonymous May 29 '17 at 14:09
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An idea (hint) for you to think about:

First, prove that for any matrix $\;A\in\Bbb R_{2\times2}\;$ , we have a decomposition:

$$A=\frac{A+A^t}2+\frac{A-A^t}2\;,\;\;\text{with}\;\;\frac{A+A^t}2\;\;\text{symmetric and}\;\;\frac{A-A^t}2\;\;\text{antisymmetric}$$

or skew-symmetric .

Show now that the set $\;\mathcal A\;$ of symmetric matrices in $\;\Bbb R_{2\times2}\;$ is a subspace of dimension $\;3\;$, the set $\;\mathcal S\;$ of antisymmetric matrices is a subspace of dimension $\;1\;$, and

$$\text{Im}\,T=\mathcal A\;,\;\;\ker T=\mathcal S$$

Martin Sleziak
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DonAntonio
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  • I can show that $dim ;\mathcal A;=3$ and $dim ;\mathcal S;=1$ considering standered basises in $\mathcal A;$ and $\mathcal S;$, respectvely. But how to conclude $\text{Im},T=\mathcal A;,;;\ker T=\mathcal S$? – user1942348 Apr 04 '17 at 10:56
  • @user1942348 What about using the dimensions theorem, also known as the rank/nullity theorem ? – DonAntonio Apr 04 '17 at 11:02
  • I can only say that rank/nullity thoerem hold good: $dim Ker{T}+\dim Im (T)=dim (;\Bbb R_{2\times2};)$ if $3+1=4$, But how to conclude ImT=A,kerT=S – user1942348 Apr 04 '17 at 11:11
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    @user I think that you should really make a deeper effort: from the above it is trivial that Im$,T\subset\mathcal A;$ ... and they both have the same finite dimension , thus .... – DonAntonio Apr 04 '17 at 12:27