When solving for eigenvector, when do you have to check every equation?

Question

For example, suppose we want to solve for the eigenvectors of:

$$A = \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} $$

We quickly find the eigenvalues are $1, -2$ i.e. $\sigma(A) = \{1, -2\}$

Then $Av_1 = \lambda_1 v_1$

$$ \begin{bmatrix} 0 & 1 \\ 2 & -1 \end{bmatrix} \begin{bmatrix}v_{11} \\ v_{12} \end{bmatrix} = \begin{bmatrix} -2v_{11} \\ -2v_{12} \end{bmatrix}$$

We get two equations:

$v_{12} = -2 v_{11}$

$2v_{11} - v_{12} = -2 v_{12}$

Solving either equation will yield equivalent eigenvectors i.e. $v_1 = [1 -2]^T$.

What account for this property? Why is it that we only need to check for one equation in this case?

Are there conditions or criteria for when we should check all equations instead just a single one?

Answer 1

Your question traces back into how one finds an eigenvector of a square matrix ${\bf{A}}$. Suppose that we are looking for nonzero vectors ${\bf{x}}$ called eigenvectors of ${\bf{A}}$ such that

$${\bf{Ax}} = \lambda {\bf{x}}\tag{1}$$

rewrite the equation in the form

$${\bf{Ax}} = \lambda {\bf{Ix}}\tag{2}$$

where ${\bf{I}}$ is the identity matrix. Now, collecting terms to the left side we can get

$$\left( {{\bf{A}} - \lambda {\bf{I}}} \right){\bf{x}} = {\bf{0}}\tag{3}$$

As you see, this is a system of linear algebraic equations. What is the requirement for this system to have nonzero solutions? Yes! the determinant of coefficient matrix should vanish which means

$$\det ({\bf{A}} - \lambda {\bf{I}}) = 0\tag{4}$$

this is the main equation that you find the eigenvalues from it. So, eigenvalues set the determinant of ${{\bf{A}} - \lambda {\bf{I}}}$ to zero. When the determinant of this matrix is zero, it implies that the equations in (3) are linearly dependent. In your example, your two equations are linearly dependent as you can easily verify that they are really the same

$$\left\{ \begin{array}{l} {v_{12}} = - 2{v_{11}}\\ 2{v_{11}} - {v_{12}} = - 2{v_{12}} \end{array} \right.\,\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,\left\{ \begin{array}{l} 2{v_{11}} + {v_{12}} = 0\\ 2{v_{11}} - {v_{12}} + 2{v_{12}} = 0 \end{array} \right.\,\,\,\, \to \,\,\,\,\,\left\{ \begin{array}{l} 2{v_{11}} + {v_{12}} = 0\\ 2{v_{11}} + {v_{12}} = 0 \end{array} \right.\tag{5}$$

In conclusion, I might say that you will never use all equations in $(2)$ or $(3)$. If ${\bf{A}}$ is a $n \times n$ matrix, then you have to use $n-1$ or less equations.

Answer 2

Let's say you start with a $n \times n$ matrix. Here $n=2$.

The defining equation for eigenvectors/eigenvalues is $Au=\lambda u$, or $(A-\lambda I)u = 0$.

This means that, given an eigenvalue $\lambda$, you are looking for the nonzero vectors $u$ such that $(A-\lambda I)u=0$. This can only happen if $A-\lambda I$ is not regular, i.e. its rank is less than $n$.

What happens then? Eigenvectors form a subspace of $\Bbb R^n$ (or $\Bbb C^n$ if you work in $\Bbb C$).

This subspace is at least of dimension $1$, which happens if $A-\lambda I$ has rank $n-1$. In that case, you will have $n-1$ equations to solve, and one "degree of freedom", that is, one parameter: hence a 1-dimension subspace.

If the rank of $A-\lambda I$ is lower, the subspace of eigenvectors associated to eigenvalue $\lambda$ will be larger, and you will have more parameters, and less equations to solve. If $\mathrm{rank} (A-\lambda I)=k$, then when solving for $(A-\lambda I)u=0$, you can solve for $k$ equations, and there remains $n-k$ parameters, or a subspace of dimension $n-k$.

If the matrix $A$ is diagonalisable, all your eigenspaces will span the whole of your linear space, but this does not always happen. It happens only if, for an eigenvalue of multiplicity $j$, the eigenspace has dimension $j$. If $j=1$, there is allways at least an eigenvector, but problems may arise when $j>1$ for some eigenvalue. Notice that the multiplicity of $\lambda$ is the power $j$ of $(t-\lambda)$ in the factorisation of the characteristic polynomial, which in turn is $\chi_A(t)=\det(A-tI_n)$. All you know for sure is that the rank of $A-\lambda I$ is less than or equal to the multiplicity of $\lambda$. It's always equal when this multiplicity is $1$. Incidentally, this also means that if a matrix has only simple eigenvalues (i.e. all have multiplicity $1$), then it's allways diagonalisable.

Hence, if $A$ is diagonalisable, the eigenvectors form a basis $P$ of your linear space, and you can write $A=P^{-1}DP$.

For instance, an upper triangular matrix with only $1$ on the diagonal can't be diagonalizable if there is any nonzero element above the diagonal. That's because when diagonalized, this matrix would necessarily be the identity (the diagonal elements are the eigenvalues). However, only the identity can yield an identity diagonalized matrix, since the equation $A=P^{-1}DP$ would yield with $D=I$, $A=P^{-1}P=I$ too.

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With $2\times2$ matrices, it's simpler: either an eigenspace has dimension $1$ (and there is thus only one equation to solve, as you noticed), either it has dimension $2$ and the initial matrix $A$ was diagonal (same reason as above).

That does not mean a $2\times2$ matrix is always diagonalisable: you still have the case where it has one eigenvalue of multiplicity $2$, but an eigenspace of dimension $1$, for instance:

$$A=\left(\begin{matrix}\lambda & 1\\ 0 & \lambda\end{matrix}\right)$$

Then there is one eigenvalue ($=\lambda$) with multiplicity $2$, and

$$A-\lambda I=\left(\begin{matrix}0 & 1\\ 0 & 0\end{matrix}\right)$$

Thus, you would have to solve for

$$\left(\begin{matrix}0 & 1\\ 0 & 0\end{matrix}\right)\left(\begin{matrix}u \\ v\end{matrix}\right)=\left(\begin{matrix}0 \\ 0\end{matrix}\right)$$

And you get the equation $v=0$. This means the eigenspace associated to $\lambda$ has one dimension, with an eigenvector $\left(\begin{matrix}1 \\ 0\end{matrix}\right)$.

Answer 3

What you do is let $\lambda = -2$ and to find the eigenvector. Since $dim(E_\lambda)=1$, there only one linear independent vector in the eigenspace you can find through two equation. Hence, you can only check one equation.

Answer 4

When you are solving for $Av=\lambda v$ and manipulate as $\lambda v-Av=0$ then you could write to $$(\lambda1\!\!1-A)v=0,\qquad (1)$$ where $1\!\!1=\left(\begin{array}{cc}1&0\\ 0&1\end{array}\right)$.

Now equation $(1)$ is an homogeneous linear system, which is known she has non trivial solutions iff the determinant of matrix $\lambda1\!\!1-A$ is zero.

This carries you to the characteristic polynomial $\chi(\lambda)$, which in case of your matrix is $\lambda^2+\lambda-2$. It is here where you can see the eigenvalues being the zeroes of that polynomial. In yours they are $$\lambda_1=-2,$$ and $$\lambda_2=1.$$

Then to find the eigenvectors you need to solve two times two equations: with $\lambda_1$ is $$\left(\begin{array}{cc}-2-0&-1\\-2&-2+1\end{array}\right) \left(\begin{array}{c}x\\y\end{array}\right)= \left(\begin{array}{c}0\\0\end{array}\right)$$ that is $$\left(\begin{array}{cc}-2&-1\\-2&-1\end{array}\right) \left(\begin{array}{c}x\\y\end{array}\right)= \left(\begin{array}{c}0\\0\end{array}\right),$$ where we can see that this correspond exactly to an only one single equation $$-2x-y=0.\qquad (2)$$ And, similarly for $\lambda_2=1$, the system should be $$\left(\begin{array}{cc}1&-1\\-2&2\end{array}\right) \left(\begin{array}{c}x\\y\end{array}\right)= \left(\begin{array}{c}0\\0\end{array}\right).$$
Here you can see two equations: $$x-y=0,\qquad (3)$$ and $$-2x+2y=0.\qquad (4)$$ But the equation $(4)$ is not very different since multiplying $(3)$ with $-2$ you get $(4)$.

This is no weird since both eigenvalues where chose to give an homogeneous linear system with matrix $\lambda1\!\!1-A$ and $\det(\lambda1\!\!1-A)=0$, meaning that its rows are linear dependent, giving you the vantage to solve only one of two equations to find the corresponding eigenvector.

Answer 5

Suppose $A$ is a $n \times n$ matrix and $\lambda$ is an eigenvalue of $A$. Then we have $$\det(A - \lambda I) = 0.$$ Therefore the system of $(A-\lambda I)v = 0$ has a nontrivial solution, which is called an eigenvector of $A$ with respect of $\lambda$. In particular, this means $$\text{nullity}(A-\lambda I) \geq 1$$ and by rank-nullity theorem, $$\text{rank}(A-\lambda I) = n - \text{nullity}(A-\lambda I) \leq n-1$$ This implies $A - \lambda I$ could have at most $n-1$ linearly independent rows, which means in solving the system $(A - \lambda I)v = 0$ in order to find an eigenvector $v$ of $A$ with respect of $\lambda$, at least $1$ row is redundant in the sense that you can write it as a linear combination of the others, and therefore you always need not consider all the $n$ rows, i.e. $n$ equations.

In particular,

1. if $A$ is $2 \times 2$, then you have only to consider $1$ equation.
2. if $A$ is, say, $3 \times 3$, then you may probably have to consider $1$ or $2$ equations.

Remark: If you know the concept of eigenspaces and multiplicity $k$ of an eigenvalue $\lambda$ of a $n \times n$ matrix $A$, then you can easily establish a sharper statement in how many equations, in terms of $n$ and $k$, are needed to determine eigenvectors, and modify the argument above accordingly.

Answer 6

Your answer is simple. If $v$ is eigenvector of an operator $A$ correspond to an eigenvalue, $cv$ is also eigenvector of operator $A$ correspond to same eigenvalue for all nonzero $c$, (actually each eigenvector introduce a subspace of domain) so for your question in 2 dimension it's not possible that your final two equations be independent and find only one vector.

Answer 7

If you have a $2\times 2$ matrix $A$ and $\lambda$ is an eigenvalue, then the eigvectors will be orthogonal to the rows of $A-\lambda I$. If $A-\lambda I$ is the $0$ matrix, then all vectors will be eigenvectors with eigenvalue $\lambda$. However, if $A-\lambda I$ is not the $0$ matrix, then the row vectors must be linearly dependent, or there could not be a vector which is orthogonal to both rows, except for the $0$ vector. That means one row is a constant multiple of the other (the constant could be $0$.) If $$ \left[\begin{array}{cc} r & s \end{array}\right] $$ is a non-zero row of $A-\lambda I$, then the eigenvector is a scalar multiple of $$ \left[\begin{array}{c}s \\ -r \end{array}\right]. $$ If both rows of $A-\lambda I$ are non-zero, then the rows are linearly dependent, and you're solving the same equation using either row, one equation being a non-zero scalar multiple of the other equation.

As an added note, if $A$ is a real $2\times 2$ symmetric matrix, then it has an orthonormal basis of eigenvectors. So, if $(A-\lambda I)x=0$ has non-trivial solution $x=\left[\begin{array}{c} u \\ v \end{array}\right]$, then $y=\left[\begin{array}{c}v \\ -u\end{array}\right]$ is automatically an eigenvector as well because $A$ must have an orthonormal basis of eigenvectors. So $2\times 2$ matrix equations are highly constrained, but especially the symmetric ones.