Eigen Vectors and Eigen Values Explained.

5 min readDec 6, 2020

Here in this blog I am going to explain what are eigen vectors and eigen values.

First we look into what is matrix multiplication will do.

Matrix Multiplication.

Let consider a matrix A of size (m x n) and matrix B of size (n x p). now we do a dot product of A.B will get a Matrix C of size (m x p). this is 2D (2 dimensional representation) in the same way we can thin of 3D,4D etc….

So, what does a dot product do actually, lets say we have a matrix A =[[3,1],[-1,3]] and then we apply a dot product with a vector X1= [1,0] if we take a dot product and represent this on a graph below. then the result vector is AX1 = [3,-1].

lets consider another vector X2 =[0,1], if we get a dot product with matrix A the the result will be AX2 = [-1,3]. which will be seen in below fig2.

So, we can think think of matrix A, which Transform a vector X1 or X2 Transformation means here matrix A rotates X1 by some angle and stretching by some factor x. This is what a matrix multiplication will do graphically.

Now we will think of a equation Ax =λx, when a dot product of vector x and A is calculated the resultant vector is a some constant time the vector x.

Now we relate our matrix multiplication with above equation which is a matrix A =[[3,1],[-1,3]] when multiplied with vector X1= [1,0] the resultant vector was AX1 = [3,-1], so this vector AX1 can’t be expressed as some constant time X1. So will find a value of vector X which satisfy the equation Ax =λx,

let vector X = [1,1] then calculate the dot product of A and X i.e A.X the resultant vector will be i.e AX = [2,2], now we can see that when dot product of A and X is some constant time X. the constant is 2. so (λ =2).

If we consider X = [1,1] only then our equation is satisfied. we can see in below equation

[[3,1],[-1,3]] . [1,1] = 2 x [1,1] = [2,2] (Ax=λx) our equation.

Here 2 is called eigen value and vector [1,1] is called eigen vector.

so, for every square matrix A there will be a vector X which when multiplied will result in a only a stretching of the original vector X by some constant value λ. so we are only interested in those values of vector x. which only stretches the original vector X. this is seen in fig 3.

So next question is how to find the value of vector X, so that when multiplied with matrix A results in a only stretching of vector X by some constant λ.

let see our equation — Ax =λx,

multiply identity matrix on each side and take the common factor. so our equation becomes-

(A-Iλ)x =0.

We have 2 case now one is

when x =0, we are not interested in this so, next case is
X ≠ 0, then determinant(A-Iλ) = 0, where (A-Iλ) is a singular matrix(Singular means determinant is zero).

lets continue with the above example only

when we substitute the values of our example with our eauation

Which results in the below matrix form.

if we take the determinant of above matrix will get

here 2,4 are the solution to the equation (A-Iλ)x =0. where λ = 2 and λ =4 are the eigen values of vector A. so we got lambda(λ) values now how to calculate value of vector x.

by substituting the value of λ in the equation 1. we get