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I have the matrix B and vector V as can be seen in the image below. Now I am trying to find out how I could show that vector v is the eigenvector of matrix B, but whenever I'm looking into eigenvector explanations I get a bit overwhelmed honestly. Could someone please show me how I would go about doing this as well as calculating the corresponding eigenvalue?

And how would I go about it the other way around? So showing that 1 is an eigenvalue of the matrix and what is the basis for its eigenspace then?

enter image description here

Thank you very much :)

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  • $\begingroup$ Are you too overwhelmed to apply the definition of an eigenvector? By (my) definition $v$ is an eigenvector of $B$ if $v\neq0$ (that's OK) and $Bv$ lies in the span of$~\{v\}$. You could start by computing $Bv$. $\endgroup$ – Marc van Leeuwen Jun 9 '17 at 10:40
  • $\begingroup$ In googling for definitions / explanations I haven't found a definition as clear as that though, what I've found is a whole lot of mathese etc that is quite confusing to me. So to get the eigenvector I just have to do Bv and that's it? How do I get the eigenvalue from that? The eigenvector here is [0,0,0] then right? $\endgroup$ – Knarf Jun 9 '17 at 10:48
  • $\begingroup$ Determining characteristic values is relatively easy. Specifically, one can rather easily prove that the statements 1) $c$ is a characteristic value of $B$ and 2) $\det(T-cI)=0$ are equivalent. Using that, you can solve for c. Finding the actual vectors is a little more involved. $\endgroup$ – Mitchell Faas Jun 9 '17 at 10:59
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    $\begingroup$ No eigenvectors are never zero (that's the $v\neq0$ part). But if $v$ is an eigenvector then $Bv$ is a scalar multiple of $v$, and that scalar, which is the eigen value, might be $0$. $\endgroup$ – Marc van Leeuwen Jun 9 '17 at 11:00
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$$B\mathbf{v}= \begin{bmatrix} 1&-2&-1\\ -4&7&3\\ 3&-4&-1 \end{bmatrix} \begin{bmatrix} 1\\ 1\\ -1 \end{bmatrix} = \begin{bmatrix} 0\\ 0\\ 0 \end{bmatrix} =0\mathbf{v} \in \mathrm{span}(\mathbf{v})$$ Thus, $\mathbf{v}$ is an eigenvector of $B$, with eigenvalue $0$.

And actually, as you know that the scalar $\lambda$ is one of the eigenvalue(s) of a matrix $A$ if and only if $\exists \mathbf{v}$ such that $$A\mathbf{v}=\lambda\mathbf{v}$$ And $\mathbf{v}$ is an eigenvector of $A$.

And also notice that the eigenvector is said to be nonzero.

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The first part of the question:

Let's think for a second about what it means to be a characteristic vector (eigenvector): Suppose $A$ a matrix and $v$ a vector, then $v$ is a characteristic vector of $A$ if their product is a multiple of that vector.

So $Av=cv$, with $c$ a constant. We call this constant $c$ the characteristic value (eigenvalue) of the vector.

Where I suspect you may be confused is that often, characteristic vectors get talked about in conjunction with linear transformations. Here it is important to understand that any linear transformation has a unique matrix associated with it that applies this transformation. So whenever you see $T(v)=cv$, you might as well read $Av=cv$.

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