Science:Math Exam Resources/Courses/MATH307/April 2005/Question 07

MATH307 April 2005

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[hide] Question 07
Determine whether or not the following two matrices are similar: $A={\begin{bmatrix}1&-1&-1\\1&3&1\\-1&-1&1\end{bmatrix}}{\text{ and }}B={\begin{bmatrix}2&0&0\\0&1&1\\0&0&2\end{bmatrix}}$

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[show] 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. We are going to work under the assumption that the two matrices are similar and use properties of similar matrices to aid us in solving this question. For two matrices to be similar $A$ and $B$ must satisfy the equation $A=P^{-1}BP$ The main properties of similar matrices which we will use to verify this equation is that: 1) Similar matrices have the same eigenvalues 2) Similar matrices have the same Jordan canonical form So, we need to get the two matrices into Jordan form and compare them. The Jordan canonical form is given by $B=GJG^{-1}$ Where $G$ is just the matrix of eigenvectors. We’re going to start with $B$ as it’s easy to pick out the eigenvalues from the diagonal, which must also be the eigenvalues of $A$ from our known properties. Eigenvalues: $\lambda =1,2,2$ So let’s get our eigenvectors! You should know how to get eigenvectors from eigenvalues (solve $Bx=\lambda x$) The eigenvectors of $B$ are: $v_{1}={\begin{bmatrix}0\\1\\0\end{bmatrix}}v_{2}={\begin{bmatrix}0\\1\\1\end{bmatrix}}v_{3}={\begin{bmatrix}1\\0\\0\end{bmatrix}}$ And so $G={\begin{bmatrix}0&0&1\\1&1&0\\0&1&0\end{bmatrix}}$ Besides knowing the formula for obtaining the inverse of a $3$x$3$ matrix, one can also obtain it by ’rref’-ing the augmented matrix: ${\begin{bmatrix}B\left.{\begin{matrix}\end{matrix}}\right|I\end{bmatrix}}$ However you want to do it, we obtain: $G^{-1}={\begin{bmatrix}0&1&-1\\0&0&1\\1&0&0\end{bmatrix}}$ Using this information we find that our J, the form we want to get, is: $J={\begin{bmatrix}1&0&0\\0&2&0\\0&0&2\end{bmatrix}}$ Now we just do the same steps again for $A$ and then we’re done! So, $A=HJH^{-1}$, and we already know our eigenvalues as they are just the same as B! The eigenvectors of $A$ are: $u_{1}={\begin{bmatrix}1\\-1\\1\end{bmatrix}}u_{2}={\begin{bmatrix}-1\\0\\1\end{bmatrix}}u_{3}={\begin{bmatrix}-1\\1\\0\end{bmatrix}}$ And so $H={\begin{bmatrix}1&-1&-1\\-1&0&1\\1&1&0\end{bmatrix}}$ and with $H$ we obtain $H^{-1}={\begin{bmatrix}1&1&1\\-1&-1&0\\1&2&1\end{bmatrix}}$ With these two matrices known we can see that: $J={\begin{bmatrix}1&0&0\\0&2&0\\0&0&2\end{bmatrix}}$ And so the two matrices have the same Jordan canonical form, and are thus similar because of the following: With us knowing that the two matrices have the same Jordan form we can show what $P$ and $P^{-1}$ will be: if $B=GJG^{-1}$ and $A=HJH^{-1}$ then $J=G^{-1}BG$ substituting that into our other equation gets: $A=HG^{-1}BGH^{-1}$ So just let $P^{-1}=HG^{-1}$ and $P=GH^{-1}$ and we get $A=P^{-1}BP$ which is what we sought out to begin with!