Learning Lab

Example 1

Find the eigenvalues of the matrix \[\begin{align*} A & =\left[\begin{array}{cc} -4 & -2\\ 3 & 3 \end{array}\right]. \end{align*}\] Solution:

The \(2\times2\) identity matrix, \[\begin{align*} I & =\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right]. \end{align*}\] Hence \[\begin{align*} A-\lambda I & =\left[\begin{array}{cc} -4 & -2\\ 3 & 3 \end{array}\right]-\lambda\left[\begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right] & \left(3\right)\\ & =\left[\begin{array}{cc} -4 & -2\\ 3 & 3 \end{array}\right]-\left[\begin{array}{cc} \lambda & 0\\ 0 & \lambda \end{array}\right] & \left(4\right)\\ & =\left[\begin{array}{cc} -4-\lambda & -2\\ 3 & 3-\lambda \end{array}\right]. & \left(5\right) \end{align*}\] Note that in practice we would not bother to include steps \(\left(3\right)\) and \(\left(4\right)\) above. We would just go directly to \(\left(5\right).\)

To find the eigenvalues we solve \(\det\left[A-\lambda I\right]=0\).33 This is the same as \[\begin{align*} \left|A-\lambda I\right| & =0. \end{align*}\] That is, \[\begin{align*} \det\left[A-\lambda I\right] & =0\\ \det\left[\begin{array}{cc} -4-\lambda & -2\\ 3 & 3-\lambda \end{array}\right] & =0\\ \left(-4-\lambda\right)\left(3-\lambda\right)-3\left(-2\right) & =0\\ -12+4\lambda-3\lambda+\lambda^{2}+6 & =0\\ \lambda^{2}+\lambda-6 & =0. \end{align*}\] The equation \(\lambda^{2}+\lambda-6=0\) is the characteristic equation of the matrix \(A.\) Factorising the characteristic equation, 44 Sometimes it is not so easy to factorise the equation. In this case you can use the quadratic formula. The solution of the quadratic \[\begin{align*} ax^{2}+bx+c & =0 \end{align*}\] is given by \[\begin{align*} x & =\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}. \end{align*}\] \[\begin{align*} 0 & =\lambda^{2}-\lambda-6\\ & =\left(\lambda+3\right)\left(\lambda-2\right) \end{align*}\] and so \(\lambda=-3\) or \(\lambda=2\). The two eigenvalues of \(A\) are \(\lambda_{1}=-3\) and \(\lambda_{2}=2\).

Example 2

Find the eigenvalues of the matrix \[\begin{align*} A & =\left[\begin{array}{cc} 2 & -1\\ 0 & 2 \end{array}\right]. \end{align*}\] Solution:

We have \[\begin{align*} A-\lambda I & =\left[\begin{array}{cc} 2-\lambda & -1\\ 0 & 2-\lambda \end{array}\right] \end{align*}\] and to find the eigenvalues we solve \(\det\left[A-\lambda I\right]=0\). That is \[\begin{align*} \det\left[\begin{array}{cc} 2-\lambda & -1\\ 0 & 2-\lambda \end{array}\right] & =0\\ \left(2-\lambda\right)\left(2-\lambda\right) & =0. \end{align*}\] So the eigenvalues of \(A\) are \(\lambda_{1}=2\) and \(\lambda_{2}=2\). Note there are two eigenvalues but they have the same value. These are called repeated eigenvalues.

Example 3

Find the eigenvectors of the matrix \[\begin{align*} A & =\left[\begin{array}{cc} -4 & -2\\ 3 & 3 \end{array}\right]. \end{align*}\]

From example 1 above we know the eigenvalues of \(A\) are \(\lambda_{1}=-3\) and \(\lambda_{2}=2\). We need to solve \[\begin{align*} \left(A-\lambda I\right)X & =0 \end{align*}\] to get the eigenvectors. Now \[\begin{align*} \left(A-\lambda I\right)\left[\begin{array}{c} x\\ y \end{array}\right] & =\left[\begin{array}{cc} -4-\lambda & -2\\ 3 & 3-\lambda \end{array}\right]\left[\begin{array}{c} x\\ y \end{array}\right] & \left(7\right)\\ & =0 \end{align*}\] where \(X=\left[\begin{array}{c} x\\ y \end{array}\right].\)

For \(\lambda_{1}=-3\) we have \[\begin{align*} \left[\begin{array}{cc} -4-\left(-3\right) & -2\\ 3 & 3-\left(-3\right) \end{array}\right]\left[\begin{array}{c} x\\ y \end{array}\right] & =0\\ \left[\begin{array}{cc} -1 & -2\\ 3 & 6 \end{array}\right]\left[\begin{array}{c} x\\ y \end{array}\right] & =0. \end{align*}\] This gives two linear equations \[\begin{align*} -x-2y & =0\\ 3x+6y & =0. \end{align*}\] These cannot be solved to give a unique solution as one is a multiple of another.55 Multiplying the first equation by \(-3\) gives the second. We have one equation and two unknowns. There will be an infinite number of solutions. Dividing the second equation by \(3\) gives \[\begin{align*} x+2y & =0\\ x & =-2y. \end{align*}\] Let \(y=t,\;t\in\mathbb{R}\) then the solution is \(x=-2t,\,y=t\) and the eigenvector has the form \[\begin{align*} \left[\begin{array}{c} x\\ y \end{array}\right] & =\left[\begin{array}{c} -2t\\ t \end{array}\right]\\ & =t\left[\begin{array}{c} -2\\ 1 \end{array}\right]. \end{align*}\] As \(t\) can be any number there are an infinite number of eigenvectors. However, the convention is that we recognise any eigenvector multiplied by a scalar is still an eigenvector and we state the eigenvector in the lowest possible terms. So, the eigenvector corresponding to \(\lambda_{1}=-3\) is \[\begin{align*} X_{1} & =\left[\begin{array}{c} -2\\ 1 \end{array}\right]. \end{align*}\]

For \(\lambda_{2}=2\), we have \[\begin{align*} \left[\begin{array}{cc} -4-2 & -2\\ 3 & 3-2 \end{array}\right]\left[\begin{array}{c} x\\ y \end{array}\right] & =0\\ \left[\begin{array}{cc} -6 & -2\\ 3 & 1 \end{array}\right]\left[\begin{array}{c} x\\ y \end{array}\right] & =0. \end{align*}\] This gives two linear equations \[\begin{align*} -6x-2y & =0\\ 3x+y & =0. \end{align*}\] Again we see that the one equation is a multiple of another.66 Multiplying the first equation by \(-3\) gives the second. We have one equation and two unknowns. There will be an infinite number of solutions. Let \(x=t\) then from the second equation, \(y=-3t\) and the eigenvector has the form \[\begin{align*} \left[\begin{array}{c} x\\ y \end{array}\right] & =\left[\begin{array}{c} t\\ -3t \end{array}\right]\\ & =t\left[\begin{array}{c} 1\\ -3 \end{array}\right]. \end{align*}\] In lowest terms, the eigenvalue corresponding to \(\lambda_{2}=2\) is \[\begin{align*} X_{2} & =\left[\begin{array}{c} 1\\ -3 \end{array}\right]. \end{align*}\]

See exercise 1 for some practice examples.

Example \(4\)

Find the eigenvalues and eigenvectors of \[\begin{align*} B & =\left[\begin{array}{ccc} 1 & 1 & -2\\ -1 & 0 & 1\\ -2 & 1 & 1 \end{array}\right]. \end{align*}\] Solution:

First find the eigenvalues by solving: \[\begin{align*} 0 & =\det\left[B-\lambda I\right]\\ & =\det\left[\begin{array}{ccc} 1-\lambda & 1 & -2\\ -1 & 0-\lambda & 1\\ -2 & 1 & 1-\lambda \end{array}\right]\\ & =\left(1-\lambda\right)\left[-\lambda\left(1-\lambda\right)-1\left(1\right)\right]-1\left[-1\left(1-\lambda\right)-1\left(-2\right)\right]-2\left[1\left(-1\right)-\left(-2\right)\left(-\lambda\right)\right]\\ & =\left(1-\lambda\right)\left(-\lambda+\lambda^{2}-1\right)-\left(-1+\lambda+2\right)-2\left(-1-2\lambda\right)\\ & =-\lambda+\lambda^{2}-1+\lambda^{2}-\lambda^{3}+\lambda+1-\lambda-2+2+4\lambda\\ & =-\lambda^{3}+2\lambda^{2}+3\lambda. \end{align*}\] The characteristic equation is: 77 Note that the characteristic equation is a cubic, that is it is a polynomial of degree \(3\). If we had a \(4\times4\) matrix, the characteristic equation would be a polynomial of degree \(4\) and so on. \[\begin{align*} -\lambda^{3}+2\lambda^{2}+3\lambda & =0. \end{align*}\] To solve this, we factorize to get: \[\begin{align*} -\lambda^{3}+2\lambda^{2}+3\lambda & =0\\ -\lambda\left(\lambda^{2}-2\lambda-3\right) & =0\\ -\lambda\left(\lambda-3\right)\left(\lambda+1\right) & =0. \end{align*}\] Hence the eigenvalues of \(B\) are: \(\lambda_{1}=0,\;\lambda_{2}=3\) and \(\lambda_{3}=-1.\) In many courses you may solve the characteristic equation using a CAS calculator. Check this with your teacher.

We now find the eigenvectors by solving equation \(\left(6\right)\) for each eigenvalue.

For \(\lambda_{1}=0,\)we have 88 Note that \[\begin{align*} X & =\left[\begin{array}{c} x\\ y\\ z \end{array}\right]. \end{align*}\] \[\begin{align*} \left(B-\lambda I\right)X & =0.\\ \left[\begin{array}{ccc} 1-0 & 1 & -2\\ -1 & 0-0 & 1\\ -2 & 1 & 1-0 \end{array}\right]\left[\begin{array}{c} x\\ y\\ z \end{array}\right] & =0\\ \left[\begin{array}{ccc} 1 & 1 & -2\\ -1 & 0 & 1\\ -2 & 1 & 1 \end{array}\right]\left[\begin{array}{c} x\\ y\\ z \end{array}\right] & =0\\ x+y-2z & =0 & \left(8\right)\\ -x+z & =0 & \left(9\right)\\ -2x+y+z & =0. & \left(10\right) \end{align*}\] Solution of these equations may be done manually or, in some courses, with a CAS calculator. Check with your teacher if you are allowed to solve them with a calculator. We solve them manually below.

From equation \(\left(9\right)\)

\[\begin{align*} z & =x. \end{align*}\] Substituting this in \(\left(8\right)\) gives \[\begin{align*} x+y-2x & =0\\ x & =y. \end{align*}\] Hence the solution is \(x=y=z=t\) where \(t\in\mathbb{R}\). Hence the eigenvector has the form \[\begin{align*} X & =t\left[\begin{array}{c} 1\\ 1\\ 1 \end{array}\right]. \end{align*}\] So the eigenvector corresponding to \(\lambda_{1}=0\) is \[\begin{align*} X_{1} & =\left[\begin{array}{c} x\\ y\\ z \end{array}\right]\\ & =\left[\begin{array}{c} 1\\ 1\\ 1 \end{array}\right]. \end{align*}\]

For \(\lambda_{2}=3\) we have \[\begin{align*} \left(B-\lambda I\right)X & =0.\\ \left[\begin{array}{ccc} 1-3 & 1 & -2\\ -1 & 0-3 & 1\\ -2 & 1 & 1-3 \end{array}\right]\left[\begin{array}{c} x\\ y\\ z \end{array}\right] & =0\\ \left[\begin{array}{ccc} -2 & 1 & -2\\ -1 & -3 & 1\\ -2 & 1 & -2 \end{array}\right]\left[\begin{array}{c} x\\ y\\ z \end{array}\right] & =0\\ -2x+y-2z & =0 & \left(11\right)\\ -x-3y+z & =0 & \left(12\right)\\ -2x+y-2z & =0. & \left(13\right) \end{align*}\] Equations \(\left(11\right)\) and \(\left(13\right)\) are identical and so we have to solve \(\left(11\right)\) and \(\left(12\right).\) Let \(z=t\) where \(t\in\mathbb{R}\) then \(\left(11\right)\) and \(\left(12\right)\) become \[\begin{align*} -2x+y & =2t & \left(14\right)\\ -x-3y & =-t. & \left(15\right) \end{align*}\] Multiplying \(\left(15\right)\) by 2 and subtracting from \(\left(14\right)\) gives \[\begin{align*} 7y & =4t\\ y & =\frac{4}{7}t. \end{align*}\] Multiplying \(\left(14\right)\) by 3 and adding to \(\left(15\right)\) gives \[\begin{align*} -7x & =5t\\ x & =-\frac{5}{7}t. \end{align*}\] Hence the eigenvector has the form \[\begin{align*} X & =t\left[\begin{array}{c} -\frac{5}{7}\\ \frac{4}{7}\\ 1 \end{array}\right]. \end{align*}\] So the eigenvector corresponding to \(\lambda_{2}=3\) is (setting \(t=7)\) \[\begin{align*} X_{2} & =\left[\begin{array}{c} x\\ y\\ z \end{array}\right]\\ & =\left[\begin{array}{c} -5\\ 4\\ 7 \end{array}\right]. & \left(16\right) \end{align*}\] Note that we chose \(t=7\) to get rid of the fraction in the eigenvector. Note also that the eigenvector \[\begin{align*} X & =\left[\begin{array}{c} 5\\ -4\\ -7 \end{array}\right] \end{align*}\] is also correct as it is just \(\left(16\right)\) multiplied by \(-1.\)

For \(\lambda_{3}=-1\) we have \[\begin{align*} \left(B-\lambda I\right)X & =0.\\ \left[\begin{array}{ccc} 1-\left(-1\right) & 1 & -2\\ -1 & 0-\left(-1\right) & 1\\ -2 & 1 & 1-\left(-1\right) \end{array}\right]\left[\begin{array}{c} x\\ y\\ z \end{array}\right] & =0\\ \left[\begin{array}{ccc} 2 & 1 & -2\\ -1 & 1 & 1\\ -2 & 1 & 2 \end{array}\right]\left[\begin{array}{c} x\\ y\\ z \end{array}\right] & =0\\ 2x+y-2z & =0 & \left(17\right)\\ -x+y+z & =0 & \left(18\right)\\ -2x+y+2z & =0. & \left(19\right) \end{align*}\] Adding \(\left(17\right)\) and \(\left(19\right)\) we get \[\begin{align*} 2y & =0\\ y & =0. \end{align*}\] Substituting this into \(\left(18\right)\) gives \[\begin{align*} -x+y+z & =0\\ x & =z. \end{align*}\] Let \(z=t,\) \(t\in\mathbb{R}\) then the eigenvector has the form \[\begin{align*} X & =\left[\begin{array}{c} x\\ y\\ z \end{array}\right]\\ & =t\left[\begin{array}{c} 1\\ 0\\ 1 \end{array}\right]. \end{align*}\] Hence the eigenvector to \(\lambda_{3}=-1\) is \[\begin{align*} X_{3} & =\left[\begin{array}{c} x\\ y\\ z \end{array}\right]\\ & =\left[\begin{array}{c} 1\\ 0\\ 1 \end{array}\right]. \end{align*}\] To summarise, the eigenvalues of \(B\) are \(\lambda_{1}=0,\;\lambda_{2}=3\) and \(\lambda_{3}=-1.\) The corresponding eigenvectors are \[\begin{align*} X_{1} & =\left[\begin{array}{c} 1\\ 1\\ 1 \end{array}\right],\;X_{2}=\left[\begin{array}{c} 5\\ -4\\ -7 \end{array}\right]\text{ and $X_{3}=\left[\begin{array}{c} 1\\ 0\\ 1 \end{array}\right]$ } \end{align*}\] respectively.

Example 5

Are \[\begin{align*} X_{1} & =\left[\begin{array}{c} 1\\ 2\\ 1 \end{array}\right]\text{ and $X_{2}=\left[\begin{array}{c} -1\\ 0\\ 1 \end{array}\right]$ } \end{align*}\] eigenvectors of \(A=\left[\begin{array}{ccc} 1 & -5 & 8\\ 1 & -2 & 1\\ 2 & -1 & -5 \end{array}\right]\)?

Solution:

One way to solve this problem is to do as in Example 4 above and calculate all the eigenvalues and their corresponding eigenvectors. However, this is the long way. A shorter and better way is to use the definition of the eigenvector in equation \(\left(1\right)\) above.99 Recall equation \(\left(1\right)\) is \[\begin{align*} AX & =\lambda X. \end{align*}\]

For \(X_{1}\) we have from the left hand side of equation \(\left(1\right)\): \[\begin{align*} AX_{1} & =\left[\begin{array}{ccc} 1 & -5 & 8\\ 1 & -2 & 1\\ 2 & -1 & -5 \end{array}\right]\left[\begin{array}{c} 1\\ 2\\ 1 \end{array}\right]\\ & =\left[\begin{array}{c} 1\times1-5\times2+8\times1\\ 1\times1-2\times2+1\times1\\ 2\times1-1\times2-5\times1 \end{array}\right]\\ & =\left[\begin{array}{c} -1\\ -2\\ -5 \end{array}\right] & \left(20\right) \end{align*}\] Now consider the right hand side of equation \(\left(1\right)\). If \(X_{1}\) is an eigenvector, there exists a number \(\lambda\) such that, from \(\left(20\right),\) \[\begin{align*} \left[\begin{array}{c} -1\\ -2\\ -5 \end{array}\right] & =\lambda\left[\begin{array}{c} 1\\ 2\\ 1 \end{array}\right]\\ & =\left[\begin{array}{c} \lambda\\ 2\lambda\\ \lambda \end{array}\right]. & \left(21\right) \end{align*}\] The first two rows of \(\left(21\right)\) imply that \(\lambda=-1.\) But then the third row implies that \(-5=-1\) which is a contradiction. We conclude that \(X_{1}\)is not an eigenvector of \(A.\)

For \(X_{2}\) we have from the left hand side of equation \(\left(1\right)\): \[\begin{align*} AX_{1} & =\left[\begin{array}{ccc} 1 & -5 & 8\\ 1 & -2 & 1\\ 2 & -1 & -5 \end{array}\right]\left[\begin{array}{c} -1\\ 0\\ 1 \end{array}\right]\\ & =\left[\begin{array}{c} 1\times\left(-1\right)-5\times0+8\times1\\ 1\times\left(-1\right)-2\times0+1\times1\\ 2\times\left(-1\right)-1\times0-5\times1 \end{array}\right]\\ & =\left[\begin{array}{c} 7\\ 0\\ -7 \end{array}\right]. & \left(22\right) \end{align*}\] Now consider the right hand side of equation \(\left(1\right)\). If \(X_{2}\) is an eigenvector, there exists a number \(\lambda\) such that, from \(\left(22\right),\) \[\begin{align*} \left[\begin{array}{c} 7\\ 0\\ -7 \end{array}\right] & =\lambda\left[\begin{array}{c} -1\\ 0\\ 1 \end{array}\right]\\ & =\left[\begin{array}{c} -\lambda\\ 0\\ \lambda \end{array}\right]. \end{align*}\] It is clear that \(\lambda=-7\) and so \(X_{2}=\left[\begin{array}{c} -1\\ 0\\ 1 \end{array}\right]\) is an eigenvector of the matrix \(A\).