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Eigenvectors of a Matrix

Tool to calculate eigenvectors of a matrix. Eigenvectors of a matrix are vectors whose direction remains unchanged after multiplying by the matrix. They are associated with an eigenvalue.

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Eigenvectors of a Matrix -

Tag(s) : Mathematics, Algebra, Symbolic Computation

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# Eigenvectors of a Matrix

## Eigenvectors Calculator

Tool to calculate eigenvectors of a matrix. Eigenvectors of a matrix are vectors whose direction remains unchanged after multiplying by the matrix. They are associated with an eigenvalue.

## Answers to Questions

### How to calculate eigen vectors of a matrix?

Consider $$M$$ a square matrix of size $$n$$ and $$\lambda_i$$ its eigenvalues. Eigenvectors are the solution of the system $$( M − \lambda I_n ) \vec{X} = \vec{0}$$ with $$I_n$$ the identity matrix.

Consider the matrix $$M=\begin{bmatrix} 1 & 2 \\ 4 & 3 \end{bmatrix}$$
Eigenvalues for the matrix $$M$$ are $$\lambda_1 = 5$$ and $$\lambda_2 = -1$$ (see tool for calculating matrices eigenvalues).
For each eigenvalue, you look for the associated eigenvector.
For $$\lambda_1 = 5$$, you can solve $$( M − 5 I_n ) X = \vec{0}$$: $$\begin{bmatrix} 1-5 & 2 \\ 4 & 3-5 \end{bmatrix} . \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$
you'll find as solution \begin{align} -4 x_1 + 2 x_2 &= 0 \\ 4 x_1 - 2 x_2 &= 0 \end{align} \iff \begin{matrix} x_1 = 1 \\ x_2 = 2 \end{matrix}
So the eigenvector associated to $$\lambda_1 = 5$$ is $$\begin{pmatrix} 1 \\ 2 \end{pmatrix}$$.
For $$\lambda_2 = -1$$, you solve $$( M + I_n ) X = \vec{0}$$ like this: \begin{bmatrix} 1+1 & 2 \\ 4 & 3+1 \end{bmatrix} . \begin{pmatrix} x_1 \\ x_2 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix} \\ \begin{align} 2 x_1 + 2 x_2 &= 0 \\ 4 x_1 + 4 x_2 &= 0 \end{align} \iff \begin{matrix} x_1 = -1 \\ x_2 = 1 \end{matrix}
So the eigenvector associated to $$\lambda_1 = -1$$ is $$\begin{pmatrix} -1 \\ 1 \end{pmatrix}$$.

### Does a zero vector as eigenvector exists?

Normally the definition of the eigenvector exclude the zero vector. However, if there are not as many independent eigenvectors as eigenvalues, dCode will display a null vector.

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## Questions / Comments

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