Finding Eigenvectors and Eigenvalues using a Determinant

Question

One possible way of finding the eigenvectors and eigenvalues (eigen-decomposition) of a matrix is to write down the characteristic equation.

$$\det(A-\lambda I)=0$$

I am interested to know in more depth why this is the case and why it works.

According to Wolfram Alpha this is due to this equation.

$$(A-\lambda I)X=0$$

where $X$ is an eigenvector.

To understand the link, what we are essentially saying here is that for scalars if we have an equation like

$$a\times x=0$$

where $x\neq 0$ then this implies that $a=0$, or if we take the modulus $\left|a\right|=0$.

For higher dimensional spaces, we have something analagous:

$$A X=0$$

implies that

$$\left|A\right|=0$$

if

$$X\neq 0$$

I know how to calculate the determinant of a matrix, but I do not know why the determinant is calculated the way it is. I suspect that there is a more in depth explanation than I currently understand. In other words, I know how to mechanically perform the operations of linear algebra, but I do not know in detail why the operations are defined the way they are.

For example, the determinant of

$$ \det \begin{pmatrix} a & b \\ c & d \\ \end{pmatrix} =ad-bc $$

I suspect this has something to do with extending the concept of an object which behaves like a scalar zero in higher dimensional spaces.

Answer 1

If I understand your question, you want to know why the solution of the equation $ det(A - \lambda I) = 0 $ are eigenvalues.

The key to understand this is there is an equivalence between the following three assumptions for each matrix A :

• (1) det(A) = 0
• (2) A is not invertible
• (3) There is a non-zero vector X such that $AX=0$

If you have this equivalence in mind, all your interrogations became clear : $$ det(A - \lambda I) = 0 \iff \exists X \ such \ that \ X \ne 0 \ and \ (A - \lambda I)X = 0 \iff AX = \lambda X $$

Now you have a second question : Why determinant is computed the way it is. The computation of the determinant have two important properties which explain its importance :

1. $det(AB) = det(A)det(B)$
2. $det(I) = 1$

So, both properties allow us to conclude that :

A is invertible $ \implies det(A) \ne 0 $ (indeed $ det(A)det(A^{-1}) = det(AA^{-1}) = det(I) = 1 $

So you have the proof $ (1) \implies (2) $ of our first set of assumptions

Thus, the computation of determinant is what it is because this computation have very good properties for studying matrices. So, when you say that something extends "the concept of an object which behaves like a scalar zero in higher dimensional spaces", actually the property which is extended is the non invertibility.

$$ \\ \\ \\ $$

Now, if you want the proofs :

• For $ det(AB) = det(A)det(B) $ you can find many proofs in https://proofwiki.org/wiki/Determinant_of_Matrix_Product or in Determinant of matrix product

• For $ det(I) = 1 $, it is very simple

• For $ (1) \implies (2) $, we showed it in the previous paragraph

• For $ (3) \implies (2) $, it is very simple : If a non-zero vector X verify $AX=0$ then A cannot be invertible because if it is the case, then : $A^{-1}AX = 0 \implies X = 0$. (But X was non-zero)

• For $ (2) \implies (3)$, it is more complex : you can prove the contraposition $ \lnot (3) \implies \lnot (2) $. If all non-zero vector X verify $ AX \ne 0 $, then we take a basis of $\mathbb{R}^n $ : $(X_1,...X_n)$, we have that $(AX_1, ..., AX_n)$ is linearly independant because otherwise we would have $ 0 = \sum_{i=0}^n \mu_i AX_i = A \sum_{i=0}^n \mu_i X_i $ and the hypothesis would not be verified. We conclude that $(AX_1, ...,AX_n)$ is a basis, so for all $ Y \in \mathbb{R}^n, \exists X \ s.t \ AX = Y$, and we can easily prove that the way we obtain X from any vector Y is linear, and so there is a matrix $A^{-1}$ such that $ AX = Y \implies Y = A^{-1}X$ . So A is invertible

• For $ (2) \implies (1) $, we can show it by using the property $MAdjugate(M) = det(M) I $ (and so we can easily compute $M^{-1}$ if $det(M) \ne 0 $, and we prove the contraposition of $(2) \implies (1) $ )

For the definition of Adjugate, see : https://en.wikipedia.org/wiki/Adjugate_matrix , and for a simple proof of the property $MAdjugate(M) = det(M) I $, see https://proofwiki.org/wiki/Matrix_Product_with_Adjugate_Matrix.

Answer 2

You have written things that are false or that do not make much sense, but the approach you are taking seems to me to go in a direction that will allow you to understand.

To help you, I'll give you an example to illustrate the link with elementary geometry. As you have done, I am happy to place myself in a two-dimensional vector space, moreover on the field of real numbers, in this case simply the plane $\mathbb R^2$.And I consider the canonical basis $\{e:=(1,0),f:=(0,1)\}$ and a linear map $$a:\mathbb R^2\to \mathbb R^2$$ $$(x,y)\mapsto(x+y,x-y)$$

Matrix of $a$ in basis $\{e:=(1,0),f:=(0,1)\}$ is $$A=\begin{bmatrix}1 & 1 \\1 & -1\end{bmatrix}$$Without going into the details of the explanation of why, $a$ is the orthogonal symmetry that you can see on the illustration composed with the homothety of ratio $\sqrt2$ For example $$D=(7,2)\to (7-2,7+2)=(5,9)=D'$$ We're going to find that again with the calculations $$det(A-\lambda I)=\begin{vmatrix}1-\lambda & 1 \\1 & -1-\lambda\end{vmatrix}=(\lambda-\sqrt{2})(\lambda+\sqrt{2})$$

The eigenvalues are then $\sqrt2$ and $-\sqrt2$.

$$a((x,y))=\sqrt2(x,y)\iff \begin{cases} x+y=\sqrt2x\\ x-y=\sqrt2y \end{cases}$$ $$\iff x(1-\sqrt2)+y=0$$ I've chosen as my associated eigenvector $u=5(1,\sqrt2-1)$

Likewise, $w=5(\sqrt2-1,-1)$

Everything is interpreted geometrically. If you think about this geometric interpretation, you understand. Otherwise, you don't understand anything.

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Note that

• in basis $\{u,v\}$, Matrix of $a$ is $$\begin{bmatrix}\sqrt{2} & 0\\0 & -\sqrt{2}\end{bmatrix}$$
• We calculate $det(A-\lambda I)$ since $$\exists \vec x\neq 0, a(\vec x)=\lambda \vec x\iff (a-\lambda id)(\vec x)=0$$$$\iff a-\lambda id \text{ is not injective}$$ $$\iff a-\lambda id \text{ is not bijective }\iff det (a-\lambda id)=0$$ $$\iff det(A-\lambda I)=0$$

Answer 3

Putting the determinant aside for a moment, let's concentrate on a homogeneous system. Let $B\in\mathbb{R}^{n\times n},x\in\mathbb{R}^n$, the following homogenous system has only the trivial solution $$ Bx=0 $$ If and only if the row reduced echolen form (rref) of $B$ is the identity matrix $I$, right?. In other words, the rank of $B$ is full (i.e. rank($B$) = $n$). For nontrivial solutions to exist, we need $\text{rank}(B) < n$. This case occurs only if $B$ is singular (i.e. $\text{rref}(B) \nrightarrow I $). Now Let's return to the eigenvalues problem. We need to find the nontrivial solutions for the following homogenous system (i.e. comes from $Ax=\lambda x \rightarrow \lambda I x - A x=0$)

$$ \underbrace{(\lambda I - A)}_{B} x = 0. $$
The matrix $(\lambda I - A)$ must be singular. "One way" to check if the square matrix is singular is $\det(B) = 0$; therefore, we have $$ \det(\lambda I - A) = 0. $$

When the eigenvalues have been computed, we can plug them back to find the eigenvectors.