Image spanned by its eigenvectos

  • Thread starter Thread starter yanky
  • Start date Start date
  • Tags Tags
    Image
AI Thread Summary
The discussion centers on whether the image of a 2x2 matrix W, defined by a mapping from R^2 to R^2, is spanned by its eigenvectors. It is noted that while the eigenvectors of W form a linearly independent set, they do not necessarily span the entire image of W. An example matrix W is provided, illustrating that its eigenvectors do not cover all of R^2, as they are multiples of a single vector. The conclusion drawn is that the image of W is not spanned by its eigenvectors, highlighting a key distinction in linear algebra. Understanding the relationship between eigenvectors and the image of a matrix is crucial for grasping linear transformations.
yanky
Messages
1
Reaction score
0
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.
 
Physics news on Phys.org
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.
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Back
Top