# Thread: Eigenvalues - Determinant proof

1. ## Eigenvalues (edit - completed)

Hey guys, I have been going around in circles for 2 hours trying to do this question. I'd really appreciate any help.

Question:

If A is a square matrix, show that:

(i) The determinant of A is equal to the product of its eigenvalues.

(ii) The trace of A is equal to the sum of its eigenvalues

edit - i have done it now. thanks.

2. Originally Posted by WWTL@WHL
Hey guys, I have been going around in circles for 2 hours trying to do this question. I'd really appreciate any help.

Question:

If A is a square matrix, show that:

(i) The determinant of A is equal to the product of its eigenvalues.

(ii) The trace of A is equal to the sum of its eigenvalues

$\displaystyle \left| \begin{array}{ccc} a & b \\ c & d \end{array} \right| = ad-bc$

Now for the characteristic equation:

$\displaystyle \left( \begin{array}{ccc} \lambda & 0 \\ 0 & \lambda \end{array} \right)- \left( \begin{array}{ccc} a & b \\ c & d \end{array} \right) = \left( \begin{array}{ccc} \lambda -a & - b \\ \lambda - c & \lambda - d \end{array} \right)$

And eigen values given by:

$\displaystyle \left| \begin{array}{ccc} \lambda -a & - b \\ - c & \lambda - d \end{array} \right| = 0$

So

$\displaystyle (\lambda - a)(\lambda - d) - (-c)(-b) = 0$

$\displaystyle \lambda^2 - a\lambda - d \lambda +ad - bc = 0$

$\displaystyle \lambda^2 +\lambda(-a - d) +ad - bc = 0$

You see where I'm going?

3. Question done.

4. Ok, I have figured it out.

No help needed now. Thanks.

5. Originally Posted by WWTL@WHL
Hey guys, I have been going around in circles for 2 hours trying to do this question. I'd really appreciate any help.

Question:

If A is a square matrix, show that:

(i) The determinant of A is equal to the product of its eigenvalues.

(ii) The trace of A is equal to the sum of its eigenvalues
Let $\displaystyle A$ be a square $\displaystyle n\times n$ matrix. Define $\displaystyle f(x) = \det (xI - A)$. This is a polynomial $\displaystyle f(x) = x^n + a_{n-1}x^{n-1} + ... + a_1 x + a_0$. If $\displaystyle x=0$ then $\displaystyle a_0 = \det (-A) \implies a_0 = (-1)^n \det (A)$. However, $\displaystyle a_0 = (-1)^n k_1k_2...k_n$ where $\displaystyle k_j$ are eigenvalues (repeated if necessary). Therefore, $\displaystyle \det (A) = k_1 ... k_n$. To prove this result for the trace that takes more work, see this.

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# prove that product of eign values of A is its determinent

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