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Kindly help me understand this statement made by my prof.

"The identity matrix I has the property that any non zero vector V is an eigenvector of eigenvalue 1."

My assumption of this statement is that the column vector (1,1) multiplied by the identity matrix is equal to the identity matrix. But the confusing part is when he says "...any non zero..". This is implying we can use other values that don't equal one. I believe the eigenvalue would change in light of the different non-1 values.

Thank you in advance.

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  • $\begingroup$ Any vector subject to the identity matrix will give you the same vector back. That is the property of the identity matrix. But in light of eigenvectors and eigenvalues, this also matches the situation for eigenvalue =1 $\endgroup$
    – imranfat
    Nov 1, 2015 at 3:15
  • $\begingroup$ part of definition. We do not include the zero vector as an eigenvector. The other matter is, for any nonzero column vector $x,$ we get $I x = x = 1 \cdot x$ $\endgroup$
    – Will Jagy
    Nov 1, 2015 at 3:15

5 Answers 5

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Let $n \geq 1$; let $\Bbb{F}$ be a field; let $I_{n}$ be the $n \times n$ identity matrix over $\Bbb{F}$. Then there is some $\lambda \in \Bbb{F}$ such that $$ \det (I_{n} - \lambda I_{n}) = (1-\lambda )^{n} = 0 $$ iff $\lambda = 1$; hence $1$ is the only eigenvalue of $I_{n}$. But $v \in \Bbb{F}^{n}$ and $v \neq 0$ only if $$ I_{n}v = v = 1\cdot v; $$ that is, every nonzero vector of $\Bbb{F}^{n}$ is an eigenvector corresponding to the eigenvalue $1$ of $I_{n}$.

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From your question, it seems that you don't understand what eigenvectors are.

If $A$ is a matrix, then we call $v$ an eigenvector if it is not zero and $Av=\lambda v$ for some constant (that is, some scalar) $\lambda$ such that $Av=\lambda v$. The constant $\lambda$ is called an eigenvalue of $A$.

Note that for every vector $v$, $Iv=1\cdot v=v$. So, if $v$ is not zero, $v$ is an eigenvector of $I$, and the associated eigenvalue is $1$.

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Eigenvectors & Eigenvalues

3Blue1Brown's video on eigenvectors and eigenvalues explains the eigenvectors and eigenvalues visually.

In general, matrix-vector multiplication $Av = b$ maps the vector $v$ to the vector $b$. Accordingly, the matrix multiplication with the identity matrix $I$ maps the vector $v$ to itself $Iv = v$.

Here is my attempt to visualize an example for $Av = b$ in 2D space:

matrix-vector multiplication example

By multiplying $A$ and $v$ we get $b$. As you can see, to get $b$ we need to stretch and rotate $v$. The rotation and scaling factor are given by $A$.

Given that $A$ is an arbitrary matrix, there can exist nonzero vectors such that when multiplied with $A$ they do not get rotated (only scaled). These vectors are eigenvectors. Therefore, we can write $\lambda v = b$ for every eigenvector $v$, where $\lambda$ is a scalar factor. We call $\lambda$ the eigenvalue.

In the following example, $A$ describes a reflection along the diagonal from bottom left to top right:

eigenvector example

Note that every vector on one of the diagonals does not get rotated, only scaled by a factor $\pm \lambda$. This means that all vectors on one of the diagonals are eigenvectors.

To answer the question

When you look at the matrix-vector multiplication with the identity matrix $Iv = v$, all vectors stay the same. In particular:

  1. There is no rotation, which means all vectors are eigenvectors (except $\vec{0}$).
  2. All vectors get scaled by a factor of $1$, which means the eigenvalue for every eigenvector is $1$.

And therefore,

The identity matrix I has the property that any non-zero vector V is an eigenvector of eigenvalue 1.

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Will Jagy's demonstration is correct. The reason for the "non-zero condition" is that for any matrix (linear function) :m, zero is a fixed point : m(o) = o, for any matrix:m.

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Every vector is an eigenvector of the identity matrix

First remember that even in the simple case where we have distinct eigenvalues and eigenvectors, then eigenvectors are not unique because if $v$ is an eigenvector, then so is every $\lambda v$ on that line.

But even if we require normal eigenvectors, this is not enough for uniqueness, because a single eigenvalue can have two distinct eigenvectors. And when that happens, any linear combination of those two eigenvectors is also an eigenvector because:

$$A(v_1 + v_2) = Av_1 + Av_2 = \lambda v_1 + \lambda v_2 = \lambda(v_1 + v_2)$$

So if for example there are two such distinct eigenvectors, then any value in that plane, and not a line, is also an eigenvector, so we could just rotate around those vectors and maintain normality with infinitely many choices.

Have a look at the concepts of:

and the key relationship:

$$1 \le \text{algebraic} \le \text{geometric} \le \text{matrix dimension}$$

So from this, we see that in the case of the identity matrix, there is a single eigenvalue of 1 with:

  • algebraic multiplicity: n
  • geometric multiplicity: n

and therefore the entire space is an eigenvector.

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