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Exercise 10.A.21

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Proof. Let $e_{1}, \dots, e_{n}$ be an orthonormal basis of $V$ with respect to which the matrix of $T$ is upper triangular. Thus we can write

M (T, (e_{1}, \dots, e_{n})) = (\begin{matrix} a_{1, 1} & \dots & a_{1, n} \\ ⋱ & ⋮ \\ 0 & a_{n, n} \end{matrix}) .

Then

| | T e_{1} | |^{2} = | a_{1, 1} |^{2} and | | T^{∗} e_{1} | |^{2} = | a_{1, 1} |^{2} + \dots + | a_{1, n} |^{2} .

Because $| | T^{∗} e_{1} | | \leq | | T e_{1} | |$ , this shows that all entries in the first row, except the first entry, are $0$ . Then we have

| | T e_{2} | |^{2} = | a_{1, 2} |^{2} + | a_{2, 2} |^{2} = | a_{2, 2} |^{2}

and

| | T^{∗} e_{2} | |^{2} = | a_{2, 2} |^{2} + \dots + | a_{2, n} |^{2} .

Because $| | T^{∗} e_{2} | | \leq | | T e_{2} | |$ , this shows that all entries in the second row, except the second entry, are $0$ . Continuing like this, we see that the matrix of $T$ is actually diagonal. Hence $e_{1}, \dots, e_{n}$ is an orthonormal basis of $V$ consisting of eigenvectors of $T$ . By the Complex Spectral Theorem (7.29) $T$ is normal. □

celiopassos

2017-10-06 00:00

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Proof. Let $e_{1}, \dots, e_{n}$ be an orthonormal basis of $V$ . By Exercise 18, we have

\begin{aligned} trace (T^{∗} T) & = | | T e_{1} | |^{2} + \dots + | | T e_{n} | |^{2} \\ \geq | | T^{∗} e_{1} | |^{2} + \dots + | | T^{∗} e_{n} | |^{2} \\ = trace (T T^{∗}) \\ = trace (T^{∗} T) . \end{aligned}

Because the first and last lines are equal, we must have equality throughout. Thus

| | T e_{1} | |^{2} + \dots + | | T e_{n} | |^{2} = | | T^{∗} e_{1} | |^{2} + \dots + | | T^{∗} e_{n} | |^{2} .

Since $| | T e_{j} | | \geq | | T^{∗} e_{j} | |$ , the equation above actually implies that $| | T e_{j} | | = | | T^{∗} e_{j} | |$ . Note that the basis $e_{1}, \dots, e_{n}$ was arbitrary and for any nonzero vector $v \in V$ , we can extend $\frac{v}{| | v | |}$ to an orthonormal basis of $V$ . Thus

| | T (\frac{v}{| | v | |}) | | = | | T^{∗} (\frac{v}{| | v | |}) | |

for all nonzero $v \in V$ . Multiplying the equation above by $| | v | |$ we get that $| | Tv | | = | | T^{∗} v | |$ . Then $T$ is normal by 7.20. □

celiopassos

2017-10-06 00:00

Exercise 10.A.21

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