# When do Eigenvectors form a basis for the space?

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?

• It's false: the null matrix has only one eigenvalue and any vector is an eigenvector, yet $\mathbf R^n$ has bases. – Bernard Apr 22 '17 at 9:41
• 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. – DonAntonio Apr 22 '17 at 9:43
• @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. – mathreadler Apr 22 '17 at 9:43
• @mathreadler Whom are you addressing? – DonAntonio Apr 22 '17 at 9:43
• I intended to address the questioner but my addressing trigger broke. – 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 $T$.

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 $u=0$".
• Yes, they take $0$ to $0$ but they may take lots of other elements to $0$ as well. – ancientmathematician Apr 22 '17 at 12:41