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i am given this matrix:

$$\begin{pmatrix} 0&1&0&-1\\ 0&0&1&0 \\ 0&0&0&0 \\ 0&0&0&0 \end{pmatrix}$$

and the basis should be these vectors: $$\begin{pmatrix} 0\\1\\0\\1 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} 1\\0\\0\\0 \end{pmatrix} $$

how come this? i am stuck, please can you give me the clue? i was looking for eigenvectors, but suddenly it says, the solution is these basis vectors. then i am stuck, so is the basis eigenvector generally?

EDIT: original matrix is this:

$$\begin{pmatrix} 2&1&1&-1\\ 0&1&-1&1 \\ 0&0&3&0 \\ 0&-2&-1&4 \end{pmatrix}$$ then i solved characteristic polynom: $det(A-\lambda E_4)$ and came to polynomial with roots 2 and 3.

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Do you mean basis of the kernel of this matrix? Basis of a matrix makes no sense. – Marc van Leeuwen Feb 15 '13 at 13:15
up vote 2 down vote accepted

It is the basis of the kernel of the matrix (when regarded as representing a linear map), that is, of the eigenspace relative to the eigenvalue zero.

Concerning the original matrix, you should be able to see that $$ \begin{pmatrix}1\\0\\0\\0\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}0\\1\\0\\1\end{pmatrix} $$ are eigenvectors with respect to the eigenvalue $2$, while $$ \begin{pmatrix}0\\0\\1\\0\end{pmatrix} \quad\text{and}\quad \begin{pmatrix}-1\\1\\0\\2\end{pmatrix} $$ are eigenvectors with respect to the eigenvalue $3$.

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@doniyor, please post the original matrix. – Andreas Caranti Feb 15 '13 at 11:55
What is 'eigenvalue formula' that you have set them into?? – Berci Feb 15 '13 at 12:22
thanks but you didnot answer my question. my question is how these two vectors come as solution? – doniyor Feb 15 '13 at 13:02
@doniyor, if $A$ is the original matrix, then the two vectors you name come out as the solutions $x$ to $A x = 2 x$. – Andreas Caranti Feb 15 '13 at 13:11
please explain me the solution depending on the matrix that i gave first. $A'x=0$ $\rightarrow$ those two vectors. why and how??? – doniyor Feb 15 '13 at 13:51

Let the matrix be $A$ and the vectors be $x,y$. Have you calculated $Ax$ and $Ay$?

Then you found that $Ax=\lambda x$ and $Ay=\mu y$ for some $\lambda,\mu$ numbers. What are these? Is there more eigenvalue?

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the eigenvalues are 2, 3. then i set eigenvalues to eigenvector formula and came to this matrix. – doniyor Feb 15 '13 at 11:53

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