Science:Math Exam Resources/Courses/MATH152/April 2012/Question 03 (c)

MATH152 April 2012

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Question 03 (c)
Let A be the matrix for the linear transformation T. Suppose we know that $\mathbf {v} _{1}={\begin{bmatrix}1\\1\\1\end{bmatrix}}$, $\mathbf {v} _{2}={\begin{bmatrix}0\\1\\1\end{bmatrix}}$, $\mathbf {v} _{3}={\begin{bmatrix}1\\0\\1\end{bmatrix}}$ are eigenvectors of A associated to the eigenvalues $\lambda =1$, $\lambda =2$, and $\lambda =3$ respectively. Find the matrix A.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
How can we use the basis vectors to build the transformation matrix?

Hint 2
Use the work from part (a) and part (b) to determine how the basis vectors are transformed using the eigenvectors.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. For any matrix A and basis vector $\mathbf {e} _{i}$ computing $A\mathbf {e} _{i}$ obtains column i of A. If A is our linear transformation matrix then computing $A\mathbf {e} _{1}$, $A\mathbf {e} _{2}$, and $A\mathbf {e} _{3}$ will give us the three columns of A and hence we'll know the matrix. From part (b), we already determined the transformation on the vector $\mathbf {e} _{1}$, $A\mathbf {e} _{1}={\begin{bmatrix}1\\-1\\-1\end{bmatrix}}$ and so this is the first column of A. To get the other relations we can use our work from part (a) where we obtained the basis vectors in terms of the eigenvectors. For $\mathbf {e} _{2}$ we have $\mathbf {e} _{2}=\mathbf {v} _{1}-\mathbf {v} _{3}$ and so $A\mathbf {e} _{2}=A\mathbf {v} _{1}-A\mathbf {v} _{3}=\mathbf {v} _{1}-3\mathbf {v} _{3}={\begin{bmatrix}-2\\1\\-2\end{bmatrix}}$ where we have used the eigenvalue/eigenvector relationships. This is the second column of A. Finally for $\mathbf {e} _{3}$ we have from part (a), $\mathbf {e} _{3}=-\mathbf {v} _{1}+\mathbf {v} _{2}+\mathbf {v} _{3}$ and so $A\mathbf {e} _{2}=-A\mathbf {v} _{1}+A\mathbf {v} _{2}+A\mathbf {v} _{3}=-\mathbf {v} _{1}+2\mathbf {v} _{2}+3\mathbf {v} _{3}={\begin{bmatrix}2\\1\\4\end{bmatrix}}$ gives us the third column of A. Therefore we have that the linear transformation matrix A is $A={\begin{bmatrix}1&-2&2\\-1&1&1\\-1&-2&4\end{bmatrix}}.$

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