# Question Video: Evaluating a Matrix Raised to the πth Power

Find [3/2, 1, and β1/2, 0]Β³β΅ and lim_(π β β) [3/2, 1, and β1/2, 0]^(π).

11:11

### Video Transcript

Find three over two, one, negative one-half, zero to the 35th power and the limit as π tends to β of three over two, one, negative one-half, zero to the πth power.

Matrix diagonalization is extremely useful for a lot of applications. And one such application is taking powers of matrices, especially when the powers are so high that taking them directly would be extremely difficult. Recall that if we have a square π-by-π matrix π΄ that is diagonalizable, then we can express π΄ as the product of π inverse π·π, where π is another π-by-π matrix and π· is a diagonal π-by-π matrix.

The columns of π are given by the eigenvectors of π΄, and the diagonal elements of π· are given by the corresponding eigenvalues of π΄. So π is the matrix formed by placing the eigenvectors of π΄ vertically in a matrix and π· is the diagonal matrix formed by placing the eigenvalues of π΄ along its diagonal. So the πΌth diagonal element of π· is the eigenvalue that corresponds with the eigenvector that makes the πΌth column of π. Expressing π΄ in this form allows us to easily raise π΄ to a power since π΄ to the π is given by π inverse π·π all to the π.

If we write this out in full, we have π inverse π·π concatenated π times. Every occurrence of π times π inverse is just equal to the identity matrix πΌ since any matrix multiplied by its inverse gives the identity. So this leaves us with π inverse times π·πΌ concatenated π minus one times π· times π. Every occurrence of π· times πΌ is just equal to π· since any matrix multiplied by the identity is just itself. So this will leave us with π΄ to the π equal to π inverse times π· to the π times π.

This right-hand side is much easier to calculate than the left-hand side since itβs very easy to take the πth power of a diagonal matrix. All we will need to do is take the πth power of each of its diagonal elements. Letβs clear some space and leave just what we need, π΄ to the π equals π inverse times π· to the π times π, where π is the square matrix whose columns are given by the eigenvectors of π΄ and π· is the diagonal matrix whose diagonal elements are given by the corresponding eigenvalues of π΄.

An eigenvector and eigenvalue is any nonzero vector π― and nonzero scalar π that satisfies the equation π΄π― equals ππ―. In other words, an eigenvector is any vector, which when operated on by π΄, is merely scaled by a scale factor of π. We can rewrite this equation first by premultiplying the vector π― on the right-hand side by the identity matrix πΌ. This does not change the value of the right-hand side but does change the equation to a purely matrix equation. We can then subtract ππΌπ― from both sides. And then factorizing gives π΄ minus ππΌ times π― is equal to the zero vector.

We already specified that the vector π― cannot be the zero vector. Therefore, the only way for this equation to hold is if the determinant of the matrix π΄ minus ππΌ is equal to zero. This gives us the characteristic polynomial the determinant of π΄ minus ππΌ is equal to zero. This is the equation that we must solve to find the eigenvalues π, which we can then substitute into the matrix equation π΄π― equals ππ― to find the eigenvectors π―.

Letβs move this equation up here for now. Applying this to our matrix here, the determinant of π΄ minus ππΌ is the determinant of three over two, one, negative one over two, zero minus π times the identity matrix one, zero, zero, one. This simplifies to the determinant of three over two minus π, one, negative one-half, negative π. Recall that for a two-by-two matrix with the elements labeled thus, the determinant is given by ππ minus ππ.

So this gives us three over two minus π times negative π minus one times negative one-half. Multiplying out the parentheses gives us π squared minus three over two π plus one-half. Setting this expression equal to zero gives us the characteristic polynomial, which is a quadratic equation in π, which we can solve using the quadratic formula. Letβs continue up here and multiply both sides of the equation by two just to make the calculations a little simpler.

Using the quadratic formula gives us π equals three plus or minus the square root of three squared minus four times two times one all over two times two. Solving this gives us our two solutions, π equals one-half or π equals one. These are the two eigenvalues of the matrix π΄. And we can substitute them back in to this equation here to find the eigenvectors π―.

Starting with our first eigenvalue π one equals one-half, we have the matrix three over two, one, negative one-half, zero times the general vector π₯, π¦ is equal to one-half times the same vector π₯, π¦. This gives us two separate equations three over two times π₯ plus π¦ equals one-half π₯ and negative one-half π₯ equals one-half π¦. These equations, however, are not distinct equations. It is possible, for instance, to rearrange the first equation to look exactly like the second. This is because there are an infinite number of corresponding eigenvectors with this eigenvalue since any scalar multiple of an eigenvector is itself an eigenvector.

Therefore, we need only parameterize one variable, say π₯, to give a simpler eigenvector as we like. If, say, we let π₯ equal one, then π¦ is equal to negative one. We, therefore, have our first eigenvector π― one equal to one, negative one, which corresponds with the first eigenvalue π one equals one-half. Letβs clear a little space before moving on to our next eigenvector.

As before we take our matrix π΄, multiply it by some general vector π₯, π¦, and make it equal to the eigenvalue, in this case, one, times the same general vector. Once again, this will give us two equations. And once again, they will not be distinct equations since there are an infinite number of eigenvectors which correspond with this eigenvalue. But this means we can just parameterize one variable, say π₯, by letting it equal some convenient value, say two, which gives π¦ equals negative one. So this gives us our second eigenvector π― two equals two, negative one, which corresponds with the second eigenvalue π two equal to one.

So we now have our two eigenvalues and the two corresponding eigenvectors. So we are ready to construct our new matrices. So the matrix π is given by listing the eigenvectors down the columns. So this gives us one, two, negative one, negative one. And our diagonal matrix π· is given by listing the corresponding eigenvalues in the same order down a diagonal. So we have π· equals one-half, zero, zero, one.

We have one more matrix we need to find, the inverse of π. Recall that for a two-by-two matrix π equal to π, π, π, π, then the inverse of π or π to the negative one is given by one over the determinant of π, which is ππ minus ππ. And then we switch the elements on the diagonal of π and switch the signs of the opposite diagonal. So this gives π, negative π, negative π, π.

So applying this to our matrix π gives us π inverse equals one over one times negative one minus two times negative one times the matrix negative one, negative two, one, one. This then simplifies to give π inverse equals negative one, negative two, one, one. We are now ready to raise π΄ to a power. And letβs start by raising it to the general power π. This will simplify things later.

So π΄ to the π equals π inverse times π· to the π times π. So we have negative one, negative two, one, one times the diagonal matrix one-half, zero, zero, one raised to the power π times one, two, negative one, negative one. When raising a diagonal matrix to a power of π, all we need to do is take the πth power of all of the diagonal entries. So for π· to the π, this gives us one-half to the π, zero, zero, one. Taking the product of the latter two matrices first, this gives us one over two to the π, two over two to the π, negative one, negative one.

Next, taking the product of these remaining two matrices gives us negative one over two to the π plus two, negative two over two to the π plus two, one over two to the π minus one, and two over two to the π minus one. Before we simplify this matrix, leaving it in this form will make it much easier to solve the second part of the question, the limit as π tends to β of this matrix to the πth power. In the limit as π tends to β, all of these terms involving one over two to the π will tend to zero. Therefore, the limit as π tends to β of three over two, one, negative one-half, zero to the πth power is just equal to two, two, negative one, negative one.

Now we can simplify this matrix by taking out a factor of one over two to the π, giving one over two to the π times two to the π plus one minus one, two to the π plus one minus two, one minus two to the π, two minus two to the π. Leaving the matrix in this form and replacing π with 35 gives us our answer. Note that these terms are very close to whole numbers and many calculators will round them.