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 $\lambda_{2}=3 ?$
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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.