How can I prove that if a linear map $T \ \in \ \text{End}(V)$ with $\dim V < \infty $ has eigenvalues ${\lambda}_{1},{\lambda}_{2}, ..., {\lambda}_{n} $, then $\text{trace}(T)$ $=$ ${\lambda}_{1}+{\lambda}_{2} + ...+{\lambda}_{n} $ and $\det(T)={\lambda}_{1}{\lambda}_{2} ...{\lambda}_{n} $?

The problem that I have is that it says that $T$ is not necessarily diagonalizable. If it is diagonalizable, then it is easy to see that there is a matrix representation with the eigenvalues in the diagonal, so this would be easy to prove. But how can I prove this with a $T$ that is not diagonalizable?

Thank you.


2 Answers 2


First of all, we will prove this result for Vector Spaces over an algebraically closed field ($\mathbb{C}$ for example).

We will prove this by induction on $\dim(V)$.

If $\dim(V) = 1$ the result is obvious.

Now, suppose that the result is valid for every vector space $W$ over an algebraically closed field with dimension $n-1$, i.e; $\forall$ $S$ $\in$ $\text{End}(W)$, $\text{trace}(S)=\lambda_1+\ldots+\lambda_{n-1}$ and $\det{(S)}=\lambda_1 \cdot \ldots\cdot \lambda_{n-1}$, where $\lambda_i$ are the eigenvalues of $S$.

Let $V$ be a vector space over an algebraically closed field $K$, such that $\dim(V)=n$ and $T$ $\in$ $\text{End}(V)$. Define $p_T: K\rightarrow K$ as the characteristic polynomial of $T$, i.e; $$p_T(t) = \det(T - tI). $$

Since $K$ is an algebraically closed field there is $\lambda$ $\in$ $K$, such that $p(\lambda) = 0$. Then there is an eigenvector $v_\lambda$ $\in$ $V$ such that $T(v_\lambda)= \lambda v_\lambda$, therefore $\exists$ $V$' a vector subespace of $V$ of dimension $n-1$, satisfying $V = \text{span}(v_\lambda)\oplus V'$, if $\beta:=\{v_1,v_2,\ldots, v_{n-1}\}$ is a basis of $V'$, then $\alpha:=\{v_\lambda,v_1,...,v_{n-1}\}$ is a basis of $V$, and

$$[T]_{\alpha}^{\alpha} = \left( \begin{array}{c|c} \lambda & \begin{array}{ccc} * & \cdots & * \end{array} \\ \hline \begin{array}{c} 0 \\ \vdots \\ 0 \end{array} & {\Huge{T'}} \end{array} \right) $$

Note that we can define a linear transformation $S: V'\rightarrow V'$ such that, $[S]_\beta^{\beta}= T'$. Then we have by induction hypothesis that $\text{trace}{(S)} = \lambda_1+... + \lambda_{n-1}$ and $\det(S) = \lambda_1\cdot \ldots\cdot\lambda_{n-1}$, where $\lambda_i$ are the eigenvalues of $S$.

Note that by our construction $\lambda, \lambda_1 ,..., \lambda_{n-1}$ are the eigenvalues of $T$ (because $p_T(t) = (\lambda - t) p_{T'} (t) $), and

$$\det(T) = \lambda\cdot \det(T') = \lambda\cdot \det([S]_{\beta}^{\beta})= \lambda\cdot \det(S) = \lambda\cdot \lambda_1\cdot \ldots\cdot\lambda_{n-1}, $$ $$\text{trace}(T) = \lambda + \text{trace}(T') = \lambda + \text{trace}([S]_{\beta}^{\beta})= \lambda + \text{trace} (S) = \lambda + \lambda_1+ \ldots+\lambda_{n-1}. $$

Which completes the demonstration if $V$ is a vector space over an algebraically closed field.

Now, if $V$ is a vector space over a field $\mathbb{F}$, and $T$ $\in$ $\text{End}_{\mathbb{F}}(V)$ we can embedding the field $\mathbb{F}$ over its algebraic closure $K$, then we can considere $T$ as an element of $\text{End}_{\mathbb{K}}(V)$. Since the eigenvalues of $T$ doesn't change if we considere $T$ as element of $\text{End}_\mathbb{F}(V)$ or element of $\text{End}_K(V)$, because the roots of $p_T(t) = \det(T - tI)$ lie in $K$, we have that $\text{trace}(T)$ = "sum of the eigenvalues of $T$" and $\det(T) = $ "product of the eigenvalues of $T$".

N.B. if you are not familiar with fields extensions, and you want to prove the result only for vector spaces over $\mathbb{R}$, you can consider $K=\mathbb{C}$ and $\mathbb{F}= \mathbb{R}$, then the demonstration will become more understandable. What we did was essentially demonstrated the result for the set of complex matrices and note that the set of real matrices is contained in the set of complex matrices, therefore the result must hold for real matrices.

  • 2
    $\begingroup$ In $[T]_{\alpha}^{\alpha}$, on the right of $\lambda$ and above $T'$, it should be stars, not zeroes. Even on an algebraically closed field, not every matrix is diagonalisable - e.g. $\left(\matrix{1&1\\0&1}\right)$. $\endgroup$ Commented May 21, 2018 at 1:26
  • $\begingroup$ You are correct, I will change, luckily this does not influence the demonstration $\endgroup$ Commented May 21, 2018 at 1:27

A matrix is triangulable as soon as its characteristic polynomial splits into linear factors, which is the case over the algebraic closure of your base field. After you write $$M=PTP^{-1}$$ where $T$ is triangular, you see that it doesn't matter that the entries of $P$ as well as the non-diagonal entries of $T$ may lie outside the base field. Only the diagonal entries of $T$ matter, and they are exactly the $\lambda_i$.

Another point of view, more pedestrian but very enlightening: think of this way of computing the determinant:

\begin{align*} \det(A) = \sum_{\tau \in S_n}\operatorname{sgn}(\tau)\,a_{1,\tau(1)}a_{2,\tau(2)} \ldots a_{n,\tau(n)}, \end{align*}

Look at the way the characteristic polynomial is computed: in $\det(M-\lambda \operatorname{Id}$), the only source of $\lambda^{n-1}$ comes from the diagonal terms. Once you have chosen $n-1$ diagonal terms, and obtained $(-\lambda)^{n-1}$, the only entry that you can select last is the $n$-th diagonal entry. But the $\lambda^{n-1}$ coefficient is also the sum of the roots of the polynomial, exactly with the sign $(-1)^{n-1}$

The same interpretation goes for the product of the roots, which is the coefficient in front of $\lambda^0$, and also the determinant of $M$.


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