I am having a bit of trouble with this proof regarding the eigenvalues. My textbook merely stated it but didn't really provide the proof:

Prove that for a matrix $A$ with eigenvalues $\lambda_{i}$: $$Coefficient(\lambda^{N-1}) = -Tr(A)$$

in the characteristic polynomial:

$$p(t) = \det(A-tI) = (-1)^n \big(t^n - (\text{tr} A) \,t^{n-1} + \dots + (-1)^n \det A\big)\,.$$

My Proof

We first consider a 3x3 matrix:

$$A-\lambda I =\begin{pmatrix} a_{11} - \lambda & a_{12} & a_{13}\\ a_{21} & a_{22} - \lambda & a_{23} \\ a_{31} & a_{32} & a _{33} - \lambda \end{pmatrix}$$

If we do a cofactor expansion around the 1st row: we obtain

$$|A - \lambda I| = (a_{11} - \lambda)(a_{22} - \lambda)(a_{33} - \lambda) + ...$$

We first note that only the $(a_{11} - \lambda)(a_{22} - \lambda)(a_{33} - \lambda)$ contributes to $\lambda ^ 2$

In fact a co-factor expansion about any row or column would result in $(a_{11} - \lambda)(a_{22} - \lambda)(a_{33} - \lambda)$ ONLY at the position of $a_{ii} - \lambda$

Expanding, we get:

$$(a_{11} - \lambda)(a_{22} - \lambda)(a_{33} - \lambda) = -\lambda^3 + (a_{11} + a_{22} + a_{33})\lambda^2 + \cdots$$

We also note that since this is a $3\times3$ matrix, the coefficient of $\lambda ^3$ (the highest power) is $-1$

We now extend this idea to an $N \times N$ matrix:


$$A-\lambda I =\begin{pmatrix} a_{11} - \lambda & a_{12} & \cdots & a_{1N}\\ a_{21} & a_{22} - \lambda &\cdots& a_{2N} \\ \vdots & \vdots & & \vdots \\ a_{N1} & a_{N2} &\cdots& a _{NN} - \lambda \end{pmatrix}$$

So using the same logic as the $3 \times 3$ matrix,

A cofactor expansion about any row or column in the $N \times N$ matrix AT THE POSITION of $a_{ii} - \lambda$ would result in the term: $$(a_{11} - \lambda)(a_{22} - \lambda)\cdots(a_{NN} - \lambda)$$ Only this term contribute to $\lambda^N-1$

In which we expect, based on the $3 \times 3$ matrix, the expansion to be:

$$(a_{11} - \lambda)(a_{22} - \lambda)\cdots(a_{NN} - \lambda) = \\ (-1)^N \lambda^N + (-1)^{N-1} (a_{11}+a_{22}+\cdots+a_{NN})\lambda^{N-1} + \cdots \\ = (-1)^N (\lambda^N -(a_{11}+a_{22}+\cdots+a_{NN})\lambda^{N-1}+\cdots) $$

thus proving that: $$Coefficient(\lambda^{N-1}) = -Tr(A)$$

Note I don't know if this is a legitimate form of proof considering that I am just extending from the idea of a $3 \times 3$ matrix

D. Soul
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  • Assuming you mean the coefficients in the characteristic polynomial, try writing the characteristic polynomial of $A$ in two ways; first as $\det(A-t I) = (-1)^n (t^n - (\text{tr} A) t^{n-1} + \dots + (-1)^n \det A)$ by forming $A - t I$ (or maybe $t I - A$, depending on what your textbook prefers) and developing the determinant along a row or column and repeating. Then also $\det(A - \lambda I) = (-1)^n(t-\lambda_1)\dots (t-\lambda_n)$. Then compare coefficients, since the two polynomials are equal. – prets May 26 '19 at 14:41
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    Possible duplicate of [Proof that the Trace of a Matrix is the sum of its Eigenvalues](https://math.stackexchange.com/questions/546155/proof-that-the-trace-of-a-matrix-is-the-sum-of-its-eigenvalues) – prets May 26 '19 at 14:46
  • hi! regarding your first part and the link that you posted. I don't really understand why the coefficient of $t^{n-1}$ is $-tr(A)$. I saw the explanation given in the link but I cant seem to make sense of it. – D. Soul May 26 '19 at 15:10
  • Try it! Make up a reasonably large matrix $A$, subtract $t I$, and compute the determinant. – prets May 26 '19 at 15:12
  • Hi I edited the question. could you comment on my proof? – D. Soul May 26 '19 at 15:57
  • That's exactly the right idea, well done! It's a fine way to prove it specifically because the result you're after really is a straight forward computation. – prets May 26 '19 at 16:05
  • @tissuepaper hey man really appreciate it. Thank you so much! – D. Soul May 26 '19 at 16:11

1 Answers1


Since $Trace(A)=a_{11}+...+a_{nn}$, and $p(t)=(t-t_1)...(t-t_n)$, where the $t_i$ are the eigenvalues of the matrix.

Now look at the coefficient of $t^{n-1}$ obtained in two possible ways:

  1. by finding the coefficient of $t^{n-1}$ using the polynomial $p(t)$. That is $-(t_1+...t_n)t^{n-1}$.

  2. by expanding $det(tI_n - A):$

$tI_n-A =\begin{pmatrix} t-a_{11} & -a_{12} & \cdots & -a_{1n}\\ -a_{21} & t-a_{22} &\cdots& -a_{2n} \\ \vdots & \vdots & & \vdots \\ -a_{n1} & -a_{n2} &\cdots& t-a_{nn} \end{pmatrix}$

Recall that $det(B) = \sum_{\sigma \in S_n} sgn(\sigma) \prod_{i=1}^{\infty} b_{i,\sigma(i)} $.

Then one of the products is $(t-a_{11})...(t-a_{nn})$, while all other products are of degree at most $(n-2)$ (so contain no $t^{n-1}$ term). therefore, the $t^{n-1}$ term of the determinant is the $t^{n-1}$ term of $(t-a_{11})...(t-a_{nn})$, which is $-(a_{11}+...a_{nn})t^{n-1}$. Now put everything together.

Locally unskillful
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