# Linear transformation of n-dimensional space has n+1 eigenvectors

There is a task:

Linear transformation of $n$-dimensional space has $n+1$ eigenvectors, such as each $n$ of them are linear independent. Find all possible matrices, that can represent such transformation.

The bold part of question made me question everything I know about eigenvectors. Assume, we have linear transformation $A$:

1. What does it mean for A to have $n$ eigenvectors?
Eigenvector is vector $x$, that satisfies $Ax=\lambda x$. But any collinear vector $cx, c \ne 0$ is also eigenvector, so it seems to me we have infinite number of vectors.
2. How can $n$-dimensional A have $n+1$ vectors?
Okay, we can make some constrain on eigenvector, like $x$ is eigenvector if $|x| = 1$, and has 'positive' direction. Not very smart, because now eigenvectors depend on norm and direction selection. Now we either have $n$ distinct eigenvalues and $n$ eigenvectors, or some repeated eigenvalue, in which case eigenvector has some free constant in it. So it also becomes many vectors. So it should be either $n$ or infinite vectors.
3. Maybe after understanding 1 and 2 I will be able to solve original problem, but any hints are welcome.

Thanks to @Catalin Zara, I can prove that this linear transformation is diagonizable!

Let $v_1 ... v_{n+1}$ be our system of vectors. $n+1$ vectors in $n$ dimensional space are linear dependent: $$v_{n+1}=a_1v_1 + ... + a_nv_n$$

Apply linear transformation A to both sides of equation, assuming that $\lambda_n$ corresponds to $v_n$ and $v_{n+1}$: $$\lambda_n v_{n+1} = \lambda_1 a_1v_1 + ... + \lambda_n a_nv_n$$

Multipling first equation by $\lambda_n$ and subtructing from second: $$0=a_1(\lambda_1-\lambda_n)v_1 + ... + a_{n-1}(\lambda_{n-1}-\lambda_n)v_{n-1}$$

This must mean one of two things: all $a_1 ... a_{n-1}$ equals zero, or $\lambda_n$ equals zero. First cannot be true, because if it is: $v_{n+1}=a_nv_n$. Thus, we have one zero eigenvalue and it means matrix A is diagonalizable.

• The problem doesn't say that there are exactly $n+1$ eigenvectors. Can you show that the transformation is diagonalizable and has exactly one eigenspace? (By the way, that was Putnam 1988 - A6). – Catalin Zara May 19 '16 at 12:58

## 1 Answer

Your reasoning at the end is not correct. First of all, your last equation should read $$0 = a_1 (\lambda_1 - \lambda_n)v_1 + \ldots + a_{n-1}(\lambda_{n-1}-\lambda_n)v_{n-1}$$ From this we conclude that $a_i (\lambda_i - \lambda_n) = 0$ for all $i$, since the $v_i$ are linearly independent. If $a_i = 0$, then we have $$0 = a_1 v_1 + \ldots + a_{i-1}v_{i-1} + a_{i+1}v_{i+1} + \ldots - v_{n+1}$$ which is impossible because $v_1 , \ldots, v_{i-1}, v_{i+1}, \ldots v_{n+1}$ are linearly independent. It follows that $\lambda_i = \lambda_n$ for all $i$, so $A$ has only one eigenvalue, hence it is a multiple of the identity matrix.

• Oh, I missed it, thank you. But statement that it holds if $\lambda_n = 0$ is valid? So it is either diagonalizable with eigenvalue 0, or multiple of identity? – DoctorMoisha May 19 '16 at 15:10
• Actually the zero matrix is a multiple of the identity too. So just "it is a multiple of the identity" will do. – Marc van Leeuwen May 19 '16 at 15:13