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- #1

- Feb 5, 2012

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Here's a question with my answer. It's pretty simple but I just want to check whether everything is perfect. Thanks in advance.

**Question:**

Let \(f:\,\mathbb{C}^2\rightarrow\mathbb{C}^2\) be a linear transformation, \(B=\{(1,0),\, (0,1)\}\) the standard basis of \(\mathbb{C}^2\) and \(A_{f,\,B}=\begin{pmatrix}3&-i\\i&3\end{pmatrix}\). Find an orthonormal basis \(C\) of engenvectors for \(f\) and \(A_{f,\,C}\).

**The eigenvectors of \(A_{f,\,B}\) in terms of the standard basis are \(v_1=(1,\, 1)\mbox{ and }v_2=(1,\,-1)\). To make this basis \(\{v_1,\,v_2\}\) orthonormal we shall divide each of the eigenvectors by their magnitudes. Hence,**

Answer:

Answer:

\[C=\left\{\left( \frac{1}{\sqrt{2}}, \, \frac{1}{\sqrt{2}} \right), \, \left(\frac{1}{\sqrt{2}} ,\, -\frac{1}{\sqrt{2}} \right) \right\}\]

Now the transformation matrix from basis \(C\) to \(B\) would be, \(\begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{pmatrix}\). It could be easily seen that the inverse of this matrix is itself. Hence the transformation matrix from basis \(B\) to \(C\) would also be the same as above. Therefore,

\[A_{f,\,C}=\begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{pmatrix}\begin{pmatrix}3&-i\\i&3\end{pmatrix}\begin{pmatrix}\frac{1}{\sqrt{2}}&\frac{1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}\end{pmatrix}=\begin{pmatrix}3&i\\-i&3\end{pmatrix}\]

Even if this is correct, I have the feeling that there should be an easier method. After all the answer is just multiplying the anti-diagonal entries of \(A_{f,\, B}\) with \(-1\).