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0,-1Find a 2 x 2 matrix A such that =A an-1 0,-2 Find nonsingular matrix P such that P-!AP is diagonal matrixHence, evaluate A2016_...

Question

0,-1Find a 2 x 2 matrix A such that =A an-1 0,-2 Find nonsingular matrix P such that P-!AP is diagonal matrixHence, evaluate A2016_

0,-1 Find a 2 x 2 matrix A such that =A an-1 0,-2 Find nonsingular matrix P such that P-!AP is diagonal matrix Hence, evaluate A2016_



Answers

Show that $A^{2}=P D^{2} P^{-1},$ where $P$ is a matrix whose columns are the eigenvectors of $A,$ and $D$ is a diagonal matrix with the corresponding eigenvalues.
$A=\left[ \begin{array}{ll}{2} & {2} \\ {0} & {1}\end{array}\right]$

In this problem we will need to use the concept of matrix multiplication. Now what we need to do is find two by two matrix E says that E square is a diagnose metrics but he is not a diagnose metrics. Now, for this consider the two way to matrix 1, 2 3 -1. Now that this matrix B equals two E. Now this matrix is not a diagonal matrix because the elements which are not on the diagonal are not equals to zero, we have two and three they are not equals to zero is not a diagonal matrix. Now if we find the product is square and e square is equal to eight times eight. So that is the matrix 123 minus one times one, 2 3 -1. So let us perform the matrix multiplication. The resulting matrix will of the product of a two by two matrix with a two by two matrix would be another two by two matrix. Now the element in the first row and first column will be the some of the products of the corresponding elements of the first row of the first matrix and the first column of the second matrix. So that means we have one times one plus two times three which is six. Now the next element which is in the first row and second column will be the some of the products of the corresponding elements in the first row of the first matrix and the second column of the second. Patrick. So that's one time too, Plus two times -1 which is -2. Similarly, the next element will be equals 23 times one and minus one times three. And the next element will be three times two plus minus one times minus one. So this is the product metrics and one plus six is seven to minus 203 minus 306 plus one is seven. So that means that the matrix e square is equal to 7007 and this is a diagonal matrix because the elements which are not on the leading diagonal are both equals to zero. So that means that the required metrics a will be 123 minus one. This is an example of 2.2 metrics, which is not a diagonal matrix, but the square, which is a diagonal matrix.

We have the matrix one minus 1 to 1. Call this matrix A and we want to show that this equals he d squared. P inverse. I'm sorry. We want to show that that's not what we want. We want to show that a squared It was p d squared d squared p inverse, where d is a diagonal matrix with the ion values of aid on the diagonal and P is a matrix with the Eiken vectors of A as the columns. Well, we see that if a equals p d p inverse then a squared multiplying by PDP in verse again gets us exactly p d squared p inverse. So it suffices to show that a is similar to a diagonal matrix, meaning that a equals p D p inverse where ideas diagonal. Because of this is the case, Then D will automatically have the Eiken values of a honest diagnose, and P will have the agon vectors of a as columns. And to show that a is similar to a diagonal matrix, we just need to show that is diagonal Izabal. And to do that, all we need to do is find to win nearly independent Aiken vectors. And to do that, we just need to find two distinct Eigen values. So let's find the wagon value or it's sufficient if we find if the alien values are distinct than the linear than the high in vectors Associate are literally dependent. So we have one minus lambda squared, plus two. This equals one minus two lambda plus Lambda squared plus two. So we get lambda squared minus two lamb duh plus three. The Eiken values are the roots of this equation. So we can find, uh, we can find the right season. The quadratic formula lambda equals two plus or minus the square root of for minus 12 All over too. Since what's under the square root here is not zero. This means that we have two distinct Eigen values, and thus we're done

We have a matrix 1404 Call it a and we want to show that a squared equals p d squared p inverse for a diagram matrix D which has the Ivan values of a on the diagonal and a matrix. A change of basis Matrix p which has the Eiken vectors of a as columns. Well, we see that if a equals p d p inverse then a squared multiplying by again gets us exactly p d squared p inverse. So all we have to show is that a is similar to a diagonal matrix D and meaning that we only have to show that this equation is true. Where d is diagonal. Because if it is, then you automatically get that the Eigen values will be on the diagonal and that the change of basis matrix people have high in vectors of a of the columns. And to do that, all you have to show is that a is diagonal Izabal to show that is diagonal Izabal It suffices to show because it is to buy to its vices to find too linearly independent. I've been vectors And to do that it suffices to have two distinct Eigen values. So let's find the Eiken values taking the determinant of a but subtracting lambda along the diagonal. So he gets one minus lambda times four minus lambda. And what do we see? We see that the roots of this equation are lambda equals one and land. It was for these air distinct, so


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