1. Solution. Because $T_\mathbb{C}$ has no real eigenvalues, if $\lambda$ is an eigenvalue of $T_\mathbb{C}$, then $\overline{\lambda}$ is also an eigenvalue of $T_\mathbb{C}$ with equal multiplicity. The determinant of $T_\mathbb{C}$, which equals the determinant of $T$, is thus a product of terms of the form $\lambda\overline{\lambda} = |\lambda|^2$, which are all greater than $0$, because $0$ is not eigenvalue of $T$. Hence $\det T > 0$.

2. Solution. By the previous exercise, $T$ has at least one eigenvalue $\alpha$, which is negative because $\det T < 0$. Suppose by contradiction that $\alpha$ is the only eigenvalue of $T$. Then this can’t be the only eigenvalue of $T_\mathbb{C}$, otherwise the $\det T$ would equal $\lambda^{\dim V}$ which is positive (because $\dim V$ is even). The other eigenvalues of $T_\mathbb{C}$ are nonreal, so they come in pairs and won’t change sign of $\det T$. It follows that $\det T$ is a product of absolute values times $\lambda$ raised to its multiplicity. Thus the multiplicity of $\lambda$ must be odd. This is a contradiction, because the sum of the multiplicities of the other eigenvalues of $T_\mathbb{C}$ is even (because they come in pairs) and the sum of all multiplicies must equal $\dim V$, which is even. Hence, our assumption that $\lambda$ was the only eigenvalue of $T$ is false.

3. Solution. (a) The characteristic polynomial of $T$ is $$ (z – \lambda_1) (z – \lambda_2) \cdots (z – \lambda_n). $$ Expanding this product we see that the coefficient of $z^{n-2}$ is the sum of the products of all pairs of distinct eigenvalues. In other words, it equals $$ \lambda_1 \sum_{j=2}^n \lambda_j + \lambda_2 \sum_{j=3}^n \lambda_j + \dots + \lambda_{n-1} \sum_{j=n}^n \lambda_j. $$ (b) It equals the sum of the products of the eigenvalues of $T$ with one term missing: $$ \sum_{j=1}^n \prod_{k \neq j} (-1)^{n-1}\lambda_k. $$ More succinctly, assuming $0$ is not an eigenvalue of $T$: $$ \sum_{j=1}^n (-1)^{n-1}\frac{\det T}{\lambda_j}. $$

4. Solution. This is follows easily from the definitions. If $\lambda$ is an eigenvalue of $T$, or $T_\mathbb{C}$, then $c\lambda$ is an eigenvalue of $cT$, or $(cT)_\mathbb{C}$, with equal multiplicity. Let $\lambda_1, \dots, \lambda_n$ denote the eigenvalues of $T_\mathbb{C}$, repeated according to its multiplicity. Then $c\lambda_1, \dots, c\lambda_n$ are the eigenvalues of $(cT)_\mathbb{C}$ and we have

$$ \det (cT) = (c \lambda_1) \cdots (c \lambda_n) = c^{\dim V}\lambda_1 \cdots \lambda_n = c^{\dim V} \det T. $$

5. Solution We give a counterexample. Let $I$ be the identity operator on $\mathbb{R}^2$. Then $$ \det(I – I) = 0 \neq 2 = \det(I) + \det(-I). $$

7. Solution. From 10.16, we have $$ \operatorname{trace} S = \operatorname{trace} A = \operatorname{trace} T $$ and, from 10.42, we have $$ \det S = \det A = \det T. $$

8. Solution. Suppose $\mathbb{F} = \mathbb{C}$. Choose an orthonormal basis of $V$ with respect to which the matrix of $T$ is upper triangular. Then, by 10.35, the determinant of $\mathcal{M}(T)$ equals the product of the entries on the diagonal. By the same reasoning used in 10.35, the determinant of $\mathcal{M}(T^*)$ is also the product of the entries on diagonal. The diagonal entries of $\mathcal{M}(T^*)$ are the conjugates of the diagonal entries of $\mathcal{M}(T)$. Hence $\det \mathcal{M}(T^*) = \overline{\det \mathcal{M}(T)}$. Now 10.42 implies that $\det T^* = \overline{\det T}$.

From 10.44, we have $$ (\det \sqrt{T^*T})^2 = \det (\sqrt{T^*T}\sqrt{T^*T}) = \det (T^*T) = \det T^* \det T = \overline{\det T} \det T = \left|\det T\right|^2. $$ Taking the square root of each side we get $\det \sqrt{T^*T} = \left|\det T\right|$.

## Chi Yuan Lau

12 Aug 2019How about 9 ?

## Linearity

13 Aug 2019Triangle inequality.