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Point) The matrixA-2has eigenvalue 1 = - 1 with an eigenspace of dimension 2_ Find a basis for the _ [ ~eigenspace:(The eigenvalues of A are ^ = -L,-1,))...

Question

Point) The matrixA-2has eigenvalue 1 = - 1 with an eigenspace of dimension 2_ Find a basis for the _ [ ~eigenspace:(The eigenvalues of A are ^ = -L,-1,))

point) The matrix A -2 has eigenvalue 1 = - 1 with an eigenspace of dimension 2_ Find a basis for the _ [ ~eigenspace: (The eigenvalues of A are ^ = -L,-1,))



Answers

The matrix $$ A=\left[\begin{array}{lll} 2 & -2 & 3 \\ 1 & -1 & 3 \\ 1 & -2 & 4 \end{array}\right] $$ has eigenvalues $\lambda_{1}=1$ and $\lambda_{2}=3$ (a) Determine a basis for the eigenspace $E_{1}$ corresponding to $\lambda_{1}=1$ and then use the GramSchmidt procedure to obtain an orthogonal basis for $E_{1}$ (b) Are the vectors in $E_{1}$ orthogonal to the vectors in $E_{2},$ the eigenspace corresponding to $\lambda_{2}=3 ?$

Problem number. Ah, 14 toe, Get the again values. We say that the determined off a minus London But the blind vanity is equal to zero. So the term mental minus longer one and one make it was born They get the one We're in a slum down minus one to to someone. That's not that These the servants equal to deal from here we together run the miners toe longer squared is equal to zero. So we can say that number is equal to zero. And And you Ah, for London equipped to and as a minus Lunda time insanity The vector is equal to zero. What you give that to me? One minor B three is equal to zero and B to minors. We three people to do you so we can say that re vector, which is 11 and one. They're worms and the bases. Ah, for the again experience anyone so that I mentioned off the organist base E one is equal to no one for London is equal to zero. Ah, and the same approach a minus Lunda Time is a anything on a definite vector is equal to you. Ah, which give that everyone minus definitely too. A plus. Two deadly three is equal to Ah is it for 20? Ah, so at that me vector is equal to buy 11 ends you plus s negative to zero on one. We're by. Yes. Wrong. It's through. Our have new vector is a print through 11 And you negative too. Do you and one bitch form the basis for the pages for Ah, the agony space e to so that I mentioned off the Internet space e to is equal to ah to so we can say that a is known defect.

This problem asks ist friend Eigen. Values of the Matrix A with the given Eigen vectors could do this using the equation. A times V is equal to Lambda Times e So first for the first Eigen vector V one, you can plug in a 4213 then V one vertically. One negative too is equal to lambda Warn times one negative, too. Then doing matrix multiplication on the left side, we got the top row to be four times one, which is four plus one times negative two, which is negative two and two times one which is two and three times negative two, which is negative. Six is equal to to negative four and that is equal toe lambda one times one negative too. So we could see the constant Lambda 11 Dividing to divide by one and negative four by the bynegative two gives us land. One is equal to two. Then for the second Eigen vector view, too, we can plug in again a 4213 Then the vector V two wishes 11 is equivalent to lambda two times 11 again matrix multiplication of the left side, The top row. We get four times what it's for. Last one times one, which is one 1212 plus three times one is three, which gives us 55 which is equivalent to land a to times 11 that if we divide five Vector 55 by the Vector 11 We find that landed too is equal to five, and those are two Eigen values.


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