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I'm studying for my exam Algebra and I'm stuck on 2 questions. Can anyone point me in the right direction? Or give me some general tips on how to solve these questions.

1) Give the function rule of $L : R^3 -> R^3$ with eigenvalues $1$, $2$ and $3$, respectivly their eigenvectores $(2, -1, 0)$, $(-1, 2, -1)$ and $(0,-1,2)$.

2) Given $L: R^3 -> R^3$ with eigenvalues $-1$, $0$ and $1$, respectivly their eigenvectores $v1$, $v2$ and $v3$. Determine $ker(L)$ and $Im(L)$.

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    The functions are defined through matrix-vector products. Consider first how this function is defined in eigen-coordinates. Seondly construct the transformation from eigen-coordinates to general coordinates if the eigenvectors are given. What is the inverse transformation? How transforms the matrix from eigen-coordinates to general coordinates? – 2017-01-06

2 Answers 2

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1) Give the function rule of $L : R^3 -> R^3$ with eigenvalues $1$, $2$ and $3$, respectivly their eigenvectores $(2, -1, 0)$, $(-1, 2, -1)$ and $(0,-1,2)$.

$L$ is completely determined by its transformation matrix $A$, the function rule is then given by: $$L: \mathbb{R^3} \to \mathbb{R^3}: x \mapsto Ax$$ You didn't get $A$, but you did get everything you need to find it since it can be diagonalized as: $$A = PDP^{-1}$$ where $D$ is a diagonal matrix with the eigenvalues on the diagonal and $P$ contains the eigenvectors in the columns.

2) Given $L: R^3 -> R^3$ with eigenvalues $-1$, $0$ and $1$, respectivly their eigenvectores $v1$, $v2$ and $v3$. Determine $ker(L)$ and $Im(L)$.

The kernel consists of all vectors that are mapped to the zero vector by $L$. Since you are given a basis of eigenvectors, these are precisely the vectors spanned by the eigenvector corresponding to eigenvalue $0$. The image is spanned by the images of the other two eigenvectors, but those images are simple multiples of those eigenvectors!

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    Thank you. So that meas that for $ker(L)$ it's $vct\{v2\}$ and for $Im(L)$ it's $vct\{v1, v3\}$? – 2017-01-06
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1) If you have a linear mapping, there is a matrix attached to it. A Matrix can be decomposed into it's eigenvalue/eigenvector decomposition. Having Eigenvectors $v^1$, $v^2$ and $v^3$ and Eigenvalues $\alpha^1$, $\alpha^2$, $\alpha^3$ of $A$, the decomposition of $A$ is $$A=\begin{bmatrix}v^1 & v^2 &v^3\end{bmatrix}\cdot \begin{bmatrix}\alpha^1&0&0\\0&\alpha^2&0\\0&0&\alpha^3\end{bmatrix} \cdot \begin{bmatrix}v^1 & v^2 &v^3\end{bmatrix}^{-1}. $$

Since you have values and vectors given, you can determine the Matrix.

2) The kernel of a matrix, is the set of vectors, that will be mapped to $0$. Think, what Eigenvalue a vector has, when mapped to 0.

Similar can be done while inspecting the image.

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    After I find A, how do I use A to find the function rule. The answer I'm supposed to find is: $ (x, y, z) →\{ x /2 − y − z /2 , x /2 + 2y − z/ 2 , x/ 2 + y + 7/ 2 z\}$ – 2017-01-06
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    You find the function rule by multication of $A$ with $\begin{bmatrix}x&y&z\end{bmatrix}^T$. By doing this, you will get a vector, the image under $A$. – 2017-01-06