# -m99.54 List possible Jordan Canonical forms for A.

• MHB
• karush
In summary: Similarly, for the second form of $A$, we have:$$A\begin{bmatrix}0\\0\\1\\0\\0\\0\end{bmatrix}=5\begin{bmatrix}0\\0\\1\\0\\0\\0\end{bmatrix}$$Therefore $\lambda=5$ is an eigenvalue with eigenvector $\mathbf v=\begin{bmatrix}0\\0\\1\\0\\0\\0\end{bmatrix}$.And for the third form of $A$, we have:$$A\begin{bmatrix}0\\0\\1\\0\\0\\0\end{bmatrix}=5\begin{bmatrix karush Gold Member MHB nmh{2000} Let A be a matrix with characteristic polynomial  p_A(t) = (t − 1)^3(t − 5)^2(t − 6) (a) List the possible Jordan Canonical forms for A. (b) Suppose all eigenspaces are one dimensional. What is the Jordan form for A in this case? Last edited:$$\begin{bmatrix} 6 \end{bmatrix}\begin{bmatrix} 5&0 \\ 0&5 \end{bmatrix}\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}\begin{bmatrix}
6&0&0&0&0&0\\
0&5&0&0&0&0\\
0&0&5&0&0&0\\
0&0&0&1&0&0\\
0&0&0&0&1&0\\
0&0&0&0&0&1
\end{bmatrix}$$Suppose 3 is an eigenvalue with (algebraic) multiplicity 4. Then the possible corresponding Jordan blocks for eigenvalue 3 are:$$\begin{bmatrix}3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix}
$$I like Serena said: Suppose 3 is an eigenvalue with (algebraic) multiplicity 4. Then the possible corresponding Jordan blocks for eigenvalue 3 are:$$\begin{bmatrix}3 & 0 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix},
\begin{bmatrix}3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix}
$$Then the possible corresponding Jordan blocks for eigenvalue 5 are: \begin{bmatrix} 5&0 \\ 0&5 \end{bmatrix} and \begin{bmatrix} 5&1 \\ 0&5 \end{bmatrix}so p_A(t) = (t − 1)^3(t − 5)^2(t − 6) has 2 possible Jordan Canonical forms for A.?what would more? karush said: Then the possible corresponding Jordan blocks for eigenvalue 5 are: \begin{bmatrix} 5&0 \\ 0&5 \end{bmatrix} and \begin{bmatrix} 5&1 \\ 0&5 \end{bmatrix}so p_A(t) = (t − 1)^3(t − 5)^2(t − 6) has 2 possible Jordan Canonical forms for A.?what would more? There are a bit more, since eigenvalue 1 also has multiple possibilities for the corresponding Jordan block. I like Serena said: There are a bit more, since eigenvalue 1 also has multiple possibilities for the corresponding Jordan block. \begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix} and \begin{bmatrix} 1&1&0\\0&1&0\\0&0&1 \end{bmatrix} and \begin{bmatrix} 1&0&0\\0&0&1\\0&0&1 \end{bmatrix} so now we can put the possible combos together? karush said: \begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix} and \begin{bmatrix} 1&1&0\\0&1&0\\0&0&1 \end{bmatrix} and \begin{bmatrix} 1&0&0\\0&0&1\\0&0&1 \end{bmatrix} so now we can put the possible combos together? I'm afraid that 3rd matrix is not correct. Instead we should have: \begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix} and \begin{bmatrix} 1&1&0\\0&1&0\\0&0&1 \end{bmatrix} and \begin{bmatrix} 1&1&0\\0&1&1\\0&0&1 \end{bmatrix}  And yes, now we can put the possible combos together. There should be 2\times 3=6 of them. I hope anyway,, I go blind looking at this... \textit{List the possible Jordan Canonical forms for A.}\\ \\ \textit{For eigenvalue 6 are}$$\begin{bmatrix} 6 \end{bmatrix}$$\textit{For eigenvalue 5 are}$$\begin{bmatrix} 5&0 \\ 0&5 \end{bmatrix},
\begin{bmatrix} 5&1 \\ 0&5 \end{bmatrix}$$\textit{ For eigenvalue 1 are}$$\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}
\begin{bmatrix} 1&1&0\\0&1&0\\0&0&1 \end{bmatrix}
\begin{bmatrix} 1&1&0\\0&1&1\\0&0&1 \end{bmatrix}$$\textit{ So the 6 possible Jordan Canonical Forms for A are:}$$\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&1\\0&0&0&0&0&1
\end{bmatrix}\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&1\\0&0&0&0&0&1
\end{bmatrix}$$(b) Suppose all eigenspaces are one dimensional. What is the Jordan form for A in this case? I suppose the answer to this is"$$\begin{bmatrix}
6\\5\\5\\1\\1\\1
\end{bmatrix}$$Last edited: karush said: I hope anyway,, I go blind looking at this... \textit{List the possible Jordan Canonical forms for A.}\\ \\ \textit{For eigenvalue 6 are}$$\begin{bmatrix} 6 \end{bmatrix}$$\textit{For eigenvalue 5 are}$$\begin{bmatrix} 5&0 \\ 0&5 \end{bmatrix},
\begin{bmatrix} 5&1 \\ 0&5 \end{bmatrix}$$\textit{ For eigenvalue 1 are}$$\begin{bmatrix} 1&0&0\\0&1&0\\0&0&1 \end{bmatrix}
\begin{bmatrix} 1&1&0\\0&1&0\\0&0&1 \end{bmatrix}
\begin{bmatrix} 1&1&0\\0&1&1\\0&0&1 \end{bmatrix}$$\textit{ So the 6 possible Jordan Canonical Forms for A are:}$$\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\begin{bmatrix}
6&0&0&0&0&0\\0&5&0&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&1\\0&0&0&0&0&1
\end{bmatrix}\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&0&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&0\\0&0&0&0&0&1
\begin{bmatrix}
6&0&0&0&0&0\\0&5&1&0&0&0\\0&0&5&0&0&0\\0&0&0&1&1&0\\0&0&0&0&1&1\\0&0&0&0&0&1
\end{bmatrix}$$Correct. karush said: (b) Suppose all eigenspaces are one dimensional. What is the Jordan form for A in this case? I suppose the answer to this is"$$\begin{bmatrix}
6\\5\\5\\1\\1\\1
\end{bmatrix}$$No, they're asking to pick one of the possible forms of A. To figure out which one, we need to know what the eigenvectors are. Can you tell what the eigenvectors are for, say, the first form of A? I like Serena said: Correct. No, they're asking to pick one of the possible forms of A. To figure out which one, we need to know what the eigenvectors are.Can you tell what the eigenvectors are for, say, the first form of A? Isn't it just 6 Actually I really don't know what to do karush said: Isn't it just 6 That is the first eigenvalue of the first form of A. An eigenvector is something different. An eigenvector \mathbf v is a non-zero vector that corresponds to an eigenvalue \lambda such that:$$A\mathbf v = \lambda \mathbf v$$In this case:$$A\begin{bmatrix}1\\0\\0\\0\\0\\0\end{bmatrix}=6\begin{bmatrix}1\\0\\0\\0\\0\\0\end{bmatrix}
Therefore $\lambda=6$ is an eigenvalue with eigenvector $\mathbf v=\begin{bmatrix}1\\0\\0\\0\\0\\0\end{bmatrix}$.

This is the only independent vector with this property, which means that the eigenspace of $6$ has dimension $1$ as requested by the problem statement.

## 1. What is a Jordan Canonical form?

A Jordan Canonical form is a special type of matrix representation that is used to simplify linear transformations in linear algebra. It is named after the mathematician Camille Jordan and is commonly used in applications such as differential equations, control theory, and quantum mechanics.

## 2. How do you determine the Jordan Canonical form for a matrix?

To determine the Jordan Canonical form for a matrix, you must first find the eigenvalues of the matrix. Then, for each eigenvalue, you must find the corresponding eigenvectors and create a Jordan block. Finally, you arrange the Jordan blocks in a specific order to form the Jordan Canonical form.

## 3. What is the significance of having multiple Jordan blocks in a Jordan Canonical form?

The number of Jordan blocks in a Jordan Canonical form represents the dimension of the eigenspace associated with each eigenvalue. This information is useful in determining the stability and other properties of a linear transformation.

## 4. Can a matrix have more than one possible Jordan Canonical form?

Yes, a matrix can have multiple possible Jordan Canonical forms. This is because there are different ways to arrange the Jordan blocks in a matrix, as long as they satisfy certain conditions such as the size and placement of the blocks.

## 5. How is the Jordan Canonical form used in real-world applications?

The Jordan Canonical form is commonly used in applications such as differential equations, control theory, and quantum mechanics. It helps simplify complex linear transformations and makes it easier to analyze and understand their behavior. It is also used in solving systems of linear equations and in finding the Jordan basis for a matrix.

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