# Find the eigenvectors corresponding to an eigenvalue

I know how to find the eigenvectors corresponding to an eigenvalue of a matrix $A$: we basically need to find the vectors of the nullspace of $\lambda I - A$, but in my case, I have a matrix $A$ like this:

$$A = \left(\begin{matrix} \cos x & -\sin x \\ \sin x & \cos x\end{matrix}\right) \left(\begin{matrix} 3 & 0 \\ 0 & 7\end{matrix}\right) \left(\begin{matrix} \cos x & \sin x \\ -\sin x & \cos x\end{matrix}\right)$$

and I cannot multiply the individual component matrices.

I have found the eigen values $\lambda = 3$ and $\lambda = 7$ for the matrix $A$, now how do I found the eigen vectors, without multiplying $A$ first?

You have $A=TDT^{-1}$. In this equation $T$ is the basis transformation matrix from the standard basis into the eigenvector basis.

When you transform from the basis $\{\vec b_1,\ldots,\vec b_n\}$ into the standard basis $\{\vec e_1,\ldots,\vec e_n\}$, then the transformation matrix has the form $(\vec b_1, \ldots \vec b_n)$ - the column vectors of the transformation matrix are exactly the basis vectors you are transforming from.

Thus the eigenvector basis are the column vectors of $T^{-1}$, i.e. $$b_1=\binom{\cos(x)}{-\sin(x)}, b_2=\binom{\sin(x)}{\cos(x)}$$

• You have it backwards if $A=T D T^{-1}$ and $D$ is diagonal then the eigenvectors of $A$ are the columns of $T$.
– Ian
Jun 5, 2015 at 21:00

Notice that the first matrix acts as a rotation of $x$ radians counterclockwise, and the third matrix acts a rotation of $x$ radians clockwise. They are inverses and cancel each other. Meaning: $A$ rotates in one direction, then do a stretch, then cancel the first rotation. Geometrically it is easy to convince your self that these matrices commute. Essentialy, $A = \begin{pmatrix} 3 & 0 \\ 0 & 7\end{pmatrix},$ and the result is obvious.

• to visualize try just multiplying a standard simple vector like $[1,1]^\top$ by a matrix with a partiular simple value of $x$ such as $\pi$ and plot the result. You should notice how it works quite quickly May 18, 2015 at 21:56
• Rotation of $x$ radians counterclockwise takes $(1,0)$ to $(\cos x, \sin x)$ and $(0,1)$ to $(-\sin x, \cos x)$. Now put that into columns. The catch is that in this case you can switch the order of the matrices. Diagonal matrices commute with everything. May 18, 2015 at 21:58
• @IvoTerek So, according to your explanation, the eigenvectors of $A$ should be $(1,0)^T$ and $(0,1)^T$? May 18, 2015 at 22:02
• Yes! With eigenvalues $3$ and $7$, in that order. May 18, 2015 at 22:03
• @IvoTerek Have a look at the eigenvectors found by tampis. May 18, 2015 at 22:05

# Eigenvector for eigenvalue 3

Given that

$$\left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = 3 \left( \begin{array}{c} 1 \\ 0 \end{array} \right),$$ we can also write $$\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = 3 \left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right),$$ or $$\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right) \left( \begin{array}{cc} \cos(x) & \sin(x) \\ -\sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = 3 \left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right),$$ or using $A$: $$A \left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = 3 \left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right),$$ Meaning the the eigenvector for the eigenvalue $3$ is given by $$\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \end{array} \right) = \left( \begin{array}{c} \cos(x) \\ \sin(x) \end{array} \right)$$

# Eigenvector for eigenvalue 7

Given that

$$\left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = 7 \left( \begin{array}{c} 0 \\ 1 \end{array} \right),$$ we can also write $$\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = 7 \left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right),$$ or $$\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{cc} 3 & 0 \\ 0 & 7 \end{array} \right) \left( \begin{array}{cc} \cos(x) & \sin(x) \\ -\sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = 7 \left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right),$$ or using $A$: $$A \left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = 3 \left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right),$$ Meaning the the eigenvector for the eigenvalue $7$ is given by $$\left( \begin{array}{cc} \cos(x) & -\sin(x) \\ \sin(x) & \cos(x) \end{array} \right) \left( \begin{array}{c} 0 \\ 1 \end{array} \right) = \left( \begin{array}{c} -\sin(x) \\ \cos(x) \end{array} \right)$$

the rotation matrix $Q = \pmatrix{\cos x & - \sin x\\ \sin x & \cos x}$ is orthogonal, that is $Q^{-1} = Q^\top.$ what you have $$A Q = Q\pmatrix{3&0\\0&7}$$ we can write the matrix equation as $$[Aq_1, Aq_2] = A[q_1, q_2] = [q_1, a_2]\pmatrix{3&0\\0&7}=[3q_1, 7q_2]$$ where $q_1, q_2$ are the columns of $Q.$ that is $$Aq_1 = 3q_1, \, Aq_2 = 7q_2 \tag 1$$

therefore the eigenvectors corresponding to the eigenvalues $3,7$ are the column vectors $q_1, q_2$ respectively.