Image spanned by its eigenvectos

[yanky]

Here's the question I'm stuck at:
Suppose a mapping from R^2 -> R^2 is defined by some 2x2 matrix W. Is the image of W spanned by the eigenvectors of W? Why or why not?

The Attempt at a Solution

I know that the eigenvectos of W for a linearly independent set. I also know that the set spanning W will consist of the smallest subspace of W consisting of linear combinations of all vectors in W. But I'm confused about what the eigenvectors actually are. I'm assuming that the answer is "yes", but I don't know why.

 

[HallsofIvy]

Suppose
$$W= \begin{bmatrix}1 & 1 \\ 0 & 1\end{bmatrix}$$
Then all eigenvectors or W are multiples of
$$\begin{bmatrix}0 \\ 1\end{bmatrix}$$
but the image of W is all or R².

 

[Architeuthis Dux]

I can provide a response to this question by first clarifying the concept of eigenvectors and then explaining the relationship between eigenvectors and the image of a matrix.

Eigenvectors are special vectors that remain unchanged in direction when multiplied by a matrix. They only change in magnitude, which is represented by a scalar value called the eigenvalue. In other words, when a matrix is multiplied by its eigenvectors, the resulting vector is a scalar multiple of the original eigenvector.

Now, let's consider the mapping from R^2 -> R^2 defined by the 2x2 matrix W. The image of this mapping is the set of all possible outputs when the matrix W is applied to any input vector in R^2. In other words, the image of W is the set of all vectors that can be obtained by multiplying W with any vector in R^2.

Since eigenvectors are unchanged in direction when multiplied by a matrix, they form a basis for the image of W. This means that any vector in the image of W can be expressed as a linear combination of the eigenvectors of W. Therefore, the image of W is spanned by its eigenvectors.

To summarize, the answer to the question is yes, the image of W is spanned by its eigenvectors. This is because eigenvectors form a basis for the image of a matrix, and any vector in the image can be expressed as a linear combination of these eigenvectors.