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The topic of this question is eigenvalues and eigenvectors.
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The question asks us to find a basis for the eigenspice of each eigenvalue of this matrix.
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And so there are multiple steps involved.
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We first want to find the characteristic equation, solve that equation to find the eigenvalues, and then use the eigenvalues to find the eigenspaces, which we describe as the set of all linear combinations of the basis vectors for those agonspaces, which are called eigenvectors.
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So let's start.
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What is the characteristic equation? well, as many of you know, the characteristic equation is the equation that solves for the roots of the characteristic polynomial, c of lambda.
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And specifically, the characteristic polynomial is the scalar function of lambda defined by the determinant of a minus lambda i, where a is the matrix we're talking about, and i is the identity matrix of the same size.
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Now, in our case, what is this going to be? well, this is what this matrix that we're taking the determinant of looks like, when we subtract lambda.
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I from a, and so we have to find its determinant.
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And now we want to solve this equation.
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So let's take a look at our characteristic polynomial here.
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3x3 determinant, and determinants are defined mathematically as co -factor expansions along any row or column, or row or column of the matrix.
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We always want to choose the row or column with the most zeros so that we have to do the least computations possible make it as easy as possible when we're calculating determinants and so this column or this row have two zeros so let's choose this column so what is our co -factor expansion it's the first entry times its co -factor plus the second entry times its co -factor plus the third times its co -factor so since zero times its co -factor so is just 0.
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All we have left is negative lambda times its cofactor.
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And the cofactor is a determinant multiplied by a multiplier of negative 1 to the power of something.
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Now, that power is the sum of the row number and column number of the entry we're looking at.
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So in the case of these entries, this sum will be even, and the multiplier will be negative 1 to the power of an even exponent.
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It is just 1, positive 1.
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The co -factor is just a determinant.
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And specifically the determinant of the sub -matrix we get by deleting the row and column, which contain our entry.
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And this is what we get.
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Now we know how to do 2x2 determinants more easily.
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It's just ad minus b .c.
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These diagonal land trees multiplied by each other, minus these ones multiplied by each other, and that gives us this, which we can expand, simplify, and factor.
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So we find the roots, or the eigenvalues of the matrix, zero, and five.
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So we can say our matrix a has these two eigenvalues.
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Now we want to find a basis for the agonspice of each eigenveillance.
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Value.
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Let's start with the second eigenvalue, lambda 2 equals 5...