Trace = Sum of eigen values - proof?

**Theju** · October 18th 2008, 06:04 PM

Hi,

I recently learnt that trace (sum of diagonal elements) of a matrix is equal to the sum of its eigen values. I was wondering why this holds true. Could anyone explain (I do not necessarily need a rigorous proof - concept by means of geometry, or what is happening in a vector space where all the columns are vectors etc would do too) ?

Thank you!

**ThePerfectHacker** · October 18th 2008, 07:19 PM

Originally Posted by Theju

Hi,

I recently learnt that trace (sum of diagonal elements) of a matrix is equal to the sum of its eigen values. I was wondering why this holds true. Could anyone explain (I do not necessarily need a rigorous proof - concept by means of geometry, or what is happening in a vector space where all the columns are vectors etc would do too) ?

Thank you!

If $A$ is a diagnolizable matrix then it is similar to a diagnol matrix consisting of its eigenvalues in the diagnols. But the trace of similar matrices is similar. Therefore the trace is the sum of its eigenvalues.

**NonCommAlg** · October 18th 2008, 08:31 PM

Originally Posted by Theju

Hi,

I recently learnt that trace (sum of diagonal elements) of a matrix is equal to the sum of its eigen values. I was wondering why this holds true. Could anyone explain (I do not necessarily need a rigorous proof - concept by means of geometry, or what is happening in a vector space where all the columns are vectors etc would do too) ?

Thank you!

we know that the sum of zeros of a polynomial $f(x)=x^n + c_1x^{n-1} + \cdots + c_n$ is $-c_1.$ now the eigenvalues of a matrix $A$ are the zeros of the polynomial $p(\lambda)=\det(\lambda I-A).$ so we only need

to prove that the coefficient of $\lambda^{n-1}$ in $p(\lambda)$ is equal to $-\text{tr}(A).$ this can be easily proved: if $A=[a_{ij}]$ is an $n \times n$ matrix, then:

$p(\lambda) = \det(\lambda I - A)= \begin{vmatrix} \lambda - a_{11} & -a_{12} & \cdots & -a_{1n} \\ -a_{21} & \lambda - a_{22} & \cdots & -a_{2n} \\ . & . & \cdots & . \\ . & . & \cdots & . \\ . & . & \cdots & . \\ -a_{n1} & -a_{n2} & \cdots & \lambda - a_{nn} \end{vmatrix}.$ it's easy to see that in the expansion of this determinant, all terms are polynomials (in $\lambda$ ) of degree at most $n-2$ except for the term

$q(\lambda)=(\lambda -a_{11})(\lambda - a_{22}) \cdots (\lambda - a_{nn}).$ thus the coefficient of $\lambda^{n-1}$ in $p(\lambda)$ is equal to the coefficient of $\lambda^{n-1}$ in $q(\lambda),$ which clearly is: $-(a_{11}+a_{22} + \cdots + a_{nn})= - \text{tr}(A). \ \ \ \Box$

**Theju** · October 20th 2008, 03:23 PM

Hi,

Thank you for your responses. They answered my questions.

**gees** · April 13th 2010, 03:13 PM

Originally Posted by NonCommAlg

we know that the sum of zeros of a polynomial $f(x)=x^n + c_1x^{n-1} + \cdots + c_n$ is $-c_1.$ now the eigenvalues of a matrix $A$ are the zeros of the polynomial $p(\lambda)=\det(\lambda I-A).$ so we only need

to prove that the coefficient of $\lambda^{n-1}$ in $p(\lambda)$ is equal to $-\text{tr}(A).$ this can be easily proved: if $A=[a_{ij}]$ is an $n \times n$ matrix, then:

$p(\lambda) = \det(\lambda I - A)= \begin{vmatrix} \lambda - a_{11} & -a_{12} & \cdots & -a_{1n} \\ -a_{21} & \lambda - a_{22} & \cdots & -a_{2n} \\ . & . & \cdots & . \\ . & . & \cdots & . \\ . & . & \cdots & . \\ -a_{n1} & -a_{n2} & \cdots & \lambda - a_{nn} \end{vmatrix}.$ it's easy to see that in the expansion of this determinant, all terms are polynomials (in $\lambda$ ) of degree at most $n-2$ except for the term

$q(\lambda)=(\lambda -a_{11})(\lambda - a_{22}) \cdots (\lambda - a_{nn}).$ thus the coefficient of $\lambda^{n-1}$ in $p(\lambda)$ is equal to the coefficient of $\lambda^{n-1}$ in $q(\lambda),$ which clearly is: $-(a_{11}+a_{22} + \cdots + a_{nn})= - \text{tr}(A). \ \ \ \Box$

**gees** · April 13th 2010, 07:57 PM

i really don't know how to solve this quest

a^2+b^2+c^2=d^2+e^2=365

**HallsofIvy** · April 14th 2010, 02:51 AM

What, exactly, is the question? You have essentially two equations, one in three unknowns, the other for two unknowns. They can't be solved for specific numbers.

By the way, you have committed three major sins here:
1) you posted a question that made no sense.

2) you posted a question in "Linear Algebra and Abstract Algebra" that has nothing to do with either.

3) you "hijacked" someone else's thread to ask a completely unrelated question.

**kstahmer** · June 3rd 2012, 12:16 AM

Let A = $\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}$

Define E = $\begin{bmatrix} \lambda I - A\end{bmatrix}$ = $\begin{bmatrix} \lambda - a_{11} & -a_{12} & \cdots & -a_{1n} \\ -a_{21} & \lambda - a_{22} & \cdots & -a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ -a_{n1} & -a_{n2} & \cdots & \lambda - a_{nn} \end{bmatrix}$ = $\begin{bmatrix} e_{11} & e_{12} & \cdots & e_{1n} \\ e_{21} & e_{22} & \cdots & e_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ e_{n1} & e_{n2} & \cdots & e_{nn} \end{bmatrix}$

Note $e_{ij}$ contains $\lambda$ , iff, $e_{ij}$ is an entry in E’s diagonal $(e_{ij}$ satisfies i = j).

Set $\chi(\lambda) = \begin{vmatrix} E \end{vmatrix} = \begin{vmatrix} \lambda I - A \end{vmatrix} = \lambda^n + c_1\lambda^{n-1} + \cdots + c_n$ . Applying the Fundamental Theorem of Algebra to $\chi(\lambda)$ repeatedly, we have $\chi(\lambda) = (\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_n)$ , where $\lambda_1, \lambda_2, \cdots \lambda_n$ are the eigenvalues of A (with possible repetitions). Multiplying factors $(\lambda - \lambda_1)(\lambda - \lambda_2) \cdots (\lambda - \lambda_n)$ and comparing the coefficients of $\lambda^{n-1}$ , we deduce $c_1 = -(\lambda_1 + \lambda_2 + \cdots + \lambda_n)$ .

Let N = {1, 2, … n} and $S_n$ denote the symmetric group of degree n $(S_n = \left\{ \sigma \ | \ \sigma: N \rightarrow N, \sigma \ \text{bijective} \right\})$ . Let $\iota$ be the identity element of $S_n$ $(\iota(1) = 1, \iota(2) = 2, \cdots \iota(n) = n)$ .

Using the Leibniz formula for determinants,

$|E| = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \ e_{1\sigma(1)}e_{2\sigma(2)} \cdots e_{n\sigma(n)} = e_{11}e_{22} \cdots e_{nn} \ + \ \sum_{\sigma \neq \iota} \text{sgn}(\sigma) \ e_{1\sigma(1)}e_{2\sigma(2)} \cdots e_{n\sigma(n)}$

For $\sigma \neq \iota$ , consider an arbitrary term $e_{1\sigma(1)}e_{2\sigma(2)} \cdots e_{n\sigma(n)}$ . There exists an $i \in N$ such that $\sigma(i) = j \ \text{with} \ j \neq i$ . Since $\sigma(i) = j$ and $\sigma$ is injective, $\sigma(j) = k \ \text{with} \ k \neq j$ . Hence, $e_{1\sigma(1)}e_{2\sigma(2)} \cdots e_{n\sigma(n)}$ contains at least two distinct non diagonal entries of E: $e_{ij}$ and $e_{jk}$ ; therefore, n - 2 is the highest power of $\lambda$ it can have.

We deduce the product of E’s diagonal terms $e_{11}e_{22} \cdots e_{nn} = (\lambda - a_{11})(\lambda - a_{22}) \cdots (\lambda - a_{nn}) = \lambda^n + d_1\lambda^{n - 1} + \cdots + d_n$ includes all $\chi(\lambda) \ \lambda^n \ \text{and} \ \lambda^{n - 1}$ terms.

As shown previously, $d_1 = -(a_{11} + a_{22} + \cdots + a_{nn})$ .

Comparing $\lambda^{n - 1}$ coefficients, we conclude:

$c_1 = d_1$

$-(\lambda_1 + \lambda_2 + \cdots + \lambda_n) = -(a_{11} + a_{22} + \cdots + a_{nn})$

$\sum_{i = 1}^n \lambda_i = tr(A)$

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Trace = Sum of eigen values - proof?

thank you for your help it has saved alot of my time, eanbling me to solve this quest

linear algebra

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