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Given $\mathbb C^2$ with the standard inner product, an operator $T(x,y) = (3x+4y, -4x+3y)$. Find $T^{*}$ and prove that $T$ is normal.

So, I took the standard basis $B = \{(1,0),(0,1)\}$ and we know it's orthogonal in respect to the standard inner product. I displayed $T$ in the $B$ basis, then transposed and conjugated it and got $[T^*]_B = \begin{bmatrix} 3 & -4 \\ 4 & 3 \end{bmatrix}$. But how exactly do we know what $T$ does on a general vector $(x,y)$ from that?

And I am also interested to know if there is another method.

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For this method to be correct, note that the basis where you take the matrix must be orthonormal, not only orthogonal. But that's the case with the one you took. Then what you did is correct. So if you managed to find the matrix of $T$, don't you know how to go backwards from the matrix of $T^*$ to deduce $T^*$'s action on $(x,y)$? –  1015 Jul 12 '13 at 16:10
    
Then $T$ is normal iff the matrices commute. But in this case, you could recognize a particular important case (skew...). That's the best method you adopted. –  1015 Jul 12 '13 at 16:14
    
Perhaps $T(x,y) = (3x - 4y, 4x + 3y)$ ? –  TheNotMe Jul 12 '13 at 16:42
    
And what do you mean by orthonormal? –  TheNotMe Jul 12 '13 at 17:00
    
Yes, that the operator $T^*$ associated with the matrix you found by transconjugation. ANd both matrices commute, so $T$ is normal. Orhtonormal set = orthogonal set of unit vectors. –  1015 Jul 13 '13 at 2:27
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1 Answer

up vote 1 down vote accepted

Note that $T^*$ is uniquely determined by the equation $<Tz,w>=<z,T^*w>$ for $z,w \in \mathbb{C^2}.$

Now let $z=(x,y)$ and $w=(a,b)$. Then $\begin{equation} <Tz,w>=<(3x+4y,-4x+3y),(a,b)> = 3x\bar{a}+4y\bar{a}-4x\bar{b}+3y\bar{b} \end{equation}$ $= <(x,y),(3a-4b,4a+3b)> = <z,T^*w>$

From the above equation it can be seen $T^*(a,b)= (3a-4b,4a+3b)$. Now just compute to check that $TT^*=T^*T$ so that $T$ becomes normal.

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Ah this is the second method I was looking for. Much appreciated! –  TheNotMe Jul 12 '13 at 17:03
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