Operations and Properties - Linear Algebra

Card 0 of 2140

Question

is an involutory matrix.

True, false, or indeterminate: 0 is an eigenvalue of .

Answer

An eigenvalue of an involutory matrix must be either 1 or . This can be seen as follows:

Let $\lambda$ be an eigenvalue of involutory matrix . Then for some eigenvector $\textbf{v}$ ,

$A\textbf{v} = \lambda \textbf{v}$

Premultiply both sides by :

$AA\textbf{v} =A \lambda \textbf{v}$

$A^{2}\textbf{v} = \lambda A \textbf{v}$

By definition, an involutory matrix has as its square, so

$I \textbf{v} = \lambda A \textbf{v}$

$\textbf{v} = \lambda A \textbf{v}$

$\frac{1}{ \lambda }\textbf{v} =A \textbf{v}$

By transitivity,

$\lambda \textbf{v} = \frac{1}{ \lambda }\textbf{v}$

Thus, $\lambda = \frac{1}{\lambda}$ , or $\lambda^{2} = 1$

It follows that $\lambda = \pm 1$ . The statement is false.

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Question

What is the dimension of the space spanned by the following vectors:

$\b{v}_1 = (1,1,0,0,0)$

$\b{v}_2 = (0,0,1,1,0)$

$\b{v}_3 = (0,0,0,0,1)$

Answer

Since there are three linearly independent vectors, they span a 3 dimensional space.

Notice that the vectors each have 5 coordinates to them. Therefore they actually span a 3 dimensional subspace of a 5 dimensional space.

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Question

In a 5 dimensional vector space, what is the maximum number of vectors you can have in a linearly dependent set?

Answer

Linearly dependent sets have no limit to the number of vectors they can have.

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Question

Consider the following set of three vectors:

$\b{v}_1,\b{v}_2,\b{v}_3$

where $\b{v}_3=2\b{v}_1+7\b{v}_2$

Is the set linearly independent?

Answer

Since $\b{v}_3$ can be written as a linear combination of of $\b{v}_1$ and $\b{v}_2$ then the set cannot be linearly independent.

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Question

Does the following row reduced echelon form of a matrix represent a linearly independent set?

$\begin{bmatrix} 1 & 0& 0& 0& \0 &0 &1 &0 & \ 0 &0 &0 &0 &\end{bmatrix}$

Answer

The set is linearly dependent because there is a row of all zeros.

Notice that having columns of all zeros does not tell if the set is linearly independent or not.

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Question

Does the following row reduced echelon form of a matrix represent a linearly independent set?

$\begin{bmatrix} 1 & 0& 0& 0& \0 &0 &1 &0 & \end{bmatrix}$

Answer

The set must be linearly independent because there are no rows of all zeros. There are columns of all zeros, but columns do not tell us if the set is linearly independent or not.

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Question

Consider the mapping $f:R^2\rightarrow R^3$ . Can f be an isomorphism?

(Hint: Think about dimension's role in isomorphism)

Answer

No, f, cannot be an isomorphism. This is because and have different dimension. Isomorphisms cannot exist between vector spaces of different dimension.

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Question

The last question showed us isomorphisms must be between vector spaces of the same dimension. This question now asks about homomorphisms.

Consider the mapping $f:R^2\rightarrow R^3$ . Can f be a homomorphism?

Answer

The answer is yes. There is no restriction on dimension for homomorphism like there is for isomorphism. Therefore f could be a homomorphism, but it is not guaranteed.

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Question

In the previous question, we said an isomorphism cannot be between vector spaces of different dimension. But are all homomorphisms between vector spaces of the same dimension an isomorphism?

Consider the homomorphism $f:R^2\rightarrow R^2$ . Is f an isomorphism?

Answer

The answer is not enough information. The reason is that it could be an isomorphism because it is between vector spaces of the same dimension, but that doesn't mean it is.

For example:

Consider the zero mapping f(x,y)= (0,0).

This mapping is not onto or 1-to-1 because all elements go to the zero vector. Therefore it is not an isomorphism even though it is a mapping between spaces with the same dimension.

Another example:

Consider the identity mapping f(x,y) = (x,y)

This is an isomorphism. It clearly preserves structure and is both onto and 1-to-1.

Thus f could be an isomorphism (example identity map) or it could NOT be an isomorphism ( Example the zero mapping)

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Question

Find the Eigen Values for Matrix .

$A=\begin{bmatrix} 5 & -3 \ 1& -2 \end{bmatrix}$

Answer

The first step into solving for eigenvalues, is adding in a $-\lambda$ along the main diagonal.

$A=\begin{bmatrix} 5-\lambda & -3 \ 1& -4-\lambda \end{bmatrix}$

Now the next step to take the determinant.

$\det(A)=(5-\lambda)(-4-\lambda)-(-3)(1)$

$\det(A)=(5-\lambda)(-4-\lambda)+3$

Now lets FOIL, and solve for $\lambda$ .

$(5-\lambda)(-4-\lambda)+3=-20-5\lambda+4\lambda+\lambda^2+3$

$=\lambda^2-\lambda-17$

Now lets use the quadratic equation to solve for $\lambda$ .

$\lambda=\frac{1\pm\sqrt{(1)^2-4(1)(-17)}}{2(1)}$

$\lambda=\frac{1\pm\sqrt{1+68}}{2}$

$\lambda=\frac{1\pm\sqrt{69}}{2}$

So our eigen values are

$\lambda=\frac{1+ \sqrt{69}}{2}$

$\lambda=\frac{1- \sqrt{69}}{2}$

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Question

Let f be a homomorphism from to . Can f be 1-to-1?

(Hint: look at the dimension of the domain and co-domain)

Answer

No, f can not be 1-to-1. The reason is because the domain has dimension 3 but the co-domain has dimension of 2. A mapping can not be 1-to-1 when the the dimension of the domain is greater than the dimension of the co-domain.

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Question

Find the eigenvalues for the matrix $\begin{bmatrix} 2 & 1\ 3& 4\end{bmatrix}$

Answer

The eigenvalues, $\lambda$ , for the matrix are values for which the determinant of $\begin{bmatrix} 2- \lambda & 1\ 3 & 4 - \lambda \end{bmatrix}$ is equal to zero. First, find the determinant:

$(2-\lambda)(4-\lambda) - (3)(1) = 8 - 4 \lambda - 2 \lambda + \lambda ^2 - 3 = \lambda ^2 - 6 \lambda + 5$

Now set the determinant equal to zero and solve this quadratic:

$\lambda^2 - 6 \lambda + 5 = 0$ this can be factored:

$(\lambda - 5 ) (\lambda - 1 ) = 0$

The eigenvalues are 5 and 1.

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Question

Which is an eigenvector for $A = \begin{bmatrix} 2 &1 \3 &4 \end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} 1\3 \end{bmatrix}$ or $\mathbf{v} = \begin{bmatrix} -3\1 \end{bmatrix}$

Answer

To determine if something is an eignevector, multiply times A:

$\begin{bmatrix} 2 &1 \3 & 4\end{bmatrix} \begin{bmatrix} 1\ 3\end{bmatrix} = \begin{bmatrix} 15\ 5\end{bmatrix}$

Since this is equivalent to $5\begin{bmatrix} 1\ 3\end{bmatrix}$ , $\begin{bmatrix} 1\ 3\end{bmatrix}$ is an eigenvector (and 5 is an eigenvalue).

$\begin{bmatrix} 2 & 1\ 3& 4\end{bmatrix} \begin{bmatrix} -3\1 \end{bmatrix} = \begin{bmatrix} -5\ -5\end{bmatrix}$

This cannot be re-written as $\mathbf{v}$ times a scalar, so this is not an eigenvector.

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Question

Find the eigenvalues for the matrix $\begin{bmatrix} -3 &6 \ 2& 1 \end{bmatrix}$

Answer

The eigenvalues are scalar quantities, $\lambda$ , where the determinant of $\begin{bmatrix} -3-\lambda & 6 \ 2& 1-\lambda \end{bmatrix}$ is equal to zero.

First, find an expression for the determinant:

$(-3-\lambda)(1-\lambda) - 12$

$=-3 - \lambda + 3 \lambda + \lambda ^2 - 12 = \lambda ^2 + 2 \lambda -15$

Now set this equal to zero, and solve:

$\lambda^2 + 2 \lambda -15 = 0$ this can be factored (or solved in another way)

$(\lambda+ 5) (\lambda-3) = 0$

The eigenvalues are -5 and 3.

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Question

Which is an eigenvector for $A = \begin{bmatrix} -3 & 6\ 2& 1\end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} -3\1 \end{bmatrix}$ or $\mathbf{v} = \begin{bmatrix} 1\ 1\end{bmatrix}$ ?

Answer

To determine if something is an eigenvector, multiply by the matrix A:

$\begin{bmatrix} -3 & 6\ 2& 1\end{bmatrix} \begin{bmatrix} -3\ 1\end{bmatrix} = \begin{bmatrix} 15\ -5 \end{bmatrix}$

This is equivalent to $-5\begin{bmatrix} -3\ 1\end{bmatrix}$ so this is an eigenvector.

$\begin{bmatrix} -3 & 6\ 2& 1\end{bmatrix} \begin{bmatrix} 1\ 1\end{bmatrix} = \begin{bmatrix} 3\ 3 \end{bmatrix}$

This is equivalent to $3\begin{bmatrix} 1\ 1\end{bmatrix}$ so this is also an eigenvector.

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Question

Often we can get information about a mapping by simply knowing the dimension of the domain and codomain.

Let f be a mapping from to . Can f be onto?

(Hint look at the dimension of the domain and codomain)

Answer

No, f cannot be onto. The reason is because the dimension of the domain (2) is less than the dimension of the codomain(3).

For a function to be onto, the dimension of the domain must be less than or equal to the dimension of the codomain.

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Question

Determine the eigenvalues for the matrix $\begin{bmatrix} -5 &2 \ 3 &-4 \end{bmatrix}$

Answer

The eigenvalues are scalar quantities $\lambda$ where the determinant of $\begin{bmatrix} -5-\lambda & 2\ 3& -4-\lambda\end{bmatrix}$ is equal to zero. First, write an expression for the determinant:

$(-5-\lambda) (-4-\lambda) -6 = 0$

$20 + 4\lambda+5 \lambda + \lambda^2 - 6 = 0$

$\lambda^2 + 9 \lambda +14 = 0$ this can be solved by factoring:

$(\lambda+2) (\lambda+7) = 0$

The solutions are -2 and -7

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Question

Which is an eigenvector for the matrix $A = \begin{bmatrix} -5 &2 \ 3& -4\end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} -2\ 3\end{bmatrix}$ or $\mathbf{v} = \begin{bmatrix} 2\ 3\end{bmatrix}$

Answer

To determine if a vector is an eigenvector, multiply with A:

$\begin{bmatrix} -5 & 2\ 3& -4 \end{bmatrix} \begin{bmatrix} -2\ 3\end{bmatrix} = \begin{bmatrix} 16\-18 \end{bmatrix}$ . This cannot be expressed as an integer times $\mathbf{u}$ , so $\mathbf{u}$ is not an eigenvector

$\begin{bmatrix} -5 & 2\ 3& -4 \end{bmatrix} \begin{bmatrix} 2\ 3\end{bmatrix} = \begin{bmatrix} -4\-6 \end{bmatrix}$ This can be expressed as $-2\begin{bmatrix} 2\ 3\end{bmatrix}$ , so $\mathbf{v}$ is an eigenvector.

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Question

$\begin{align}&\text{Find any and all eigenvalues for the matrix }\&A=\begin{bmatrix}6&-13\8&-19\end{bmatrix}\end{align}$

Answer

$\begin{align}&\text{Eigenvalues of a matrix, typically noted as }\lambda\text{, are those values which satisfy}\&det(A-\lambda I)\text{(I being the identity matrix). For the matrix }A=\begin{bmatrix}6&-13\8&-19\end{bmatrix}\&\text{We can represent that equation as follows }\begin{bmatrix}6 - \lambda &-13\8&- \lambda - 19\end{bmatrix} \&\text{For a square matrix with dimensions of }2\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&\text{From that, we can find the eigenvalues:}\&13\lambda + \lambda ^{2} - 10\&\lambda_{1}=0.73;\lambda_{2}=-13.73\end{align}$

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Question

$\begin{align}&\text{Calculate the eigenvalues for the matrix }\&\begin{bmatrix}0&-2\6&9\end{bmatrix}\end{align}$

Answer

$\begin{align}&\text{Eigenvalues of a matrix A, often denoted as }\lambda\text{, are values which satisfy}\&det(A-\lambda I)\text{, where I represents the identity matrix. For the given matrix }\begin{bmatrix}0&-2\6&9\end{bmatrix}\&\text{We can write a matrix of the form }\begin{bmatrix}-\lambda&-2\6&9 - \lambda\end{bmatrix}\&\text{For a square matrix with dimensions of }2\&det\begin{vmatrix} a&b \ c&d \end{vmatrix}=ad-bc\&\text{Knowing this, we can expand and solve:}\&\lambda ^{2} - 9\lambda + 12\&\lambda_{1}=1.63;\lambda_{2}=7.37\end{align}$

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