I have a linear operator $T:\mathbb{R}^n\to\mathbb{R}^n$ defined by $Tu=Au $ where $A$ is a matrix defined as $A^2=βA$, $β$ is a constant. I'm trying to show that if $β=0,$ $A≠0$, then there cant exist a basis of Eigenvectors in $\mathbb{R}^n$. Here is what I've tried:

Assume that $β=0$ is an eigenvalue so for every $u∈\mathbb{R}^n$ we have $Tu=βu=0$, since $T$ is linear $Tu=0$ if and only if $u$ is $0$ but $0$ can't be an eigenvector, so there can't be a basis of eigenvectors when $β=0$.

Did I go about that right?

  • $\begingroup$ It's false: the null matrix has only one eigenvalue and any vector is an eigenvector, yet $\mathbf R^n$ has bases. $\endgroup$ – Bernard Apr 22 '17 at 9:41
  • $\begingroup$ Again this question? This was already asked yesterday and already yesterday I told the asker (and perhaps some others, too. I can't remember) that the zero operator is diagonalizable and any basis of $\;\Bbb R^n\;$ is a basis of eigenvectors of that operator...weird. $\endgroup$ – DonAntonio Apr 22 '17 at 9:43
  • $\begingroup$ @user The eigenvectors of an operator don't always span the whole space. If one algebraic multiplicity differs from geometric multiplicity then there will be subspaces which aren't spanned by eigenvectors. $\endgroup$ – mathreadler Apr 22 '17 at 9:43
  • $\begingroup$ @mathreadler Whom are you addressing? $\endgroup$ – DonAntonio Apr 22 '17 at 9:43
  • $\begingroup$ I intended to address the questioner but my addressing trigger broke. $\endgroup$ – mathreadler Apr 22 '17 at 9:45

If $\beta=0$, $A$ is a nilpotent matrix, and the only diagonalisable nilpotent matrix is the null matrix.


"Did I go about that right?" No, you are very confused indeed. But I will try to help.

For a start, your hypotheses include $\beta=0$. So get rid of $\beta$, it's just a red herring!

Then you say "where $A$ is a matrix defined by $A^2=\beta A$". As I say, get rid of the irrelevant $\beta$ and this would read "where $A$ is a matrix defined by $A^2=0$". This makes no sense, this does not define $A$. You surely mean "where $A$ is such that $A^2=0$. Clarity helps.

Looking ahead a bit, you'll see there's another hypothesis, $A\not=0$. So get it upfront with the others!

So your hypothesis is that $T:\mathbb{R}^n\to\mathbb{R}^n$ is the linear transformation defined by $T(u)=Au$, where $A$ is a matrix such that $A^2=0$ and $A\not=0$.

Your problem is to prove there is not a basis of eigenvectors of $T$.

Now some comments on your answer.

You write "$0$ is an eigenvalue so for every $u\in\mathbb{R}^n$ we have $Tu=0$.

This is fatally wrong, could scarcely be worse. The definition of "eigenstuff" tells us the correct statement is "$0$ is an eigenvalue if for some $u\in\mathbb{R}^n$ we have $Tu=0$ and $u\not=0$."

You the write "Since $T$ is linear $Tu=0$ if and only if $u=0$".

This is fatally wrong. There are many linear maps which kill non-zero vectors.

The other solution tells you how to solve the underlying question. I've tried to answer your question about whether your approach is right. It's not, and you need to try to understand why.

  • $\begingroup$ thank you for clearing that for me but when you said "There are many linear maps which kill non-zero vectors" ,now i'm more confused cause i was depending on the theorem that Linear Transformations Take Zero to Zero. $\endgroup$ – user3133165 Apr 22 '17 at 11:47
  • $\begingroup$ Yes, they take $0$ to $0$ but they may take lots of other elements to $0$ as well. $\endgroup$ – ancientmathematician Apr 22 '17 at 12:41

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