Operations and Properties - Linear Algebra

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Question

A linear map $f:R^4 \rightarrow R^2$ has a null space that is spanned by the vectors,

and . is this function 1-to-1?

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Answer

This linear mapping is not 1-to-1. We know this because the null space is spanned by two vectors. Therefore the null space has dimension . A linear map is 1-to-1 only if it has a null space with dimension zero. This linear map doesn't have a null space of dimension zero, therefore it is not 1-to-1.

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Question

A linear map $f:R^4 \rightarrow R^3$ has a rank of 3. Is the linear map, , 1-to-1?

(Hint- Use the formula where

is the dimension of the domain

is the dimension of the null space

is the rank of the linear map )

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Answer

The answer is no because the dimension of the null space is not zero. This comes from the equation

We know that the domain is which has dimension of . Therefore

Also from the problem statement .

Plugging these into the equation gives

Since the dimension of the null space is and not , then the function can't be 1-to-1.

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Question

$\begin{bmatrix} 2&5 \ 4& 5\ 3&2 \ 0&1 \end{bmatrix}^{T}=$

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Answer

Transposing a matrix simply means to make the columns of the original matrix the rows in the transposed matrix. Example: $\begin{bmatrix} 1& 2&3 \ 4&5 &6 \ 7&8 &9 \end{bmatrix}^{T} = \begin{bmatrix} 1&4 &7 \ 2&5 &8 \ 3&6 &9 \end{bmatrix}$ ie. column 1 become row 1, column 2 becomes row 2, etc.

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Question

A linear map $f:R^4 \rightarrow R^5$ has a rank of 4. Is the linear map, , 1-to-1?

(Hint- Use the formula where

is the dimension of the domain

is the dimension of the null space

is the rank of the linear map )

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Answer

The answer is yes because the dimension of the domain and the rank are equal. This implies that the dimension of the null space is zero.

This comes from the equation

We know that the domain is which has dimension of . Therefore

Also from the problem statement .

Plugging these into the equation gives

Since the dimension of the null space is then the function is 1-to-1.

Notice that whenever , then is always zero. Thus whenever , the linear mapping is 1-to-1.

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Question

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

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Answer

Transposing a matrix simply means to make the columns of the original matrix the rows in the transposed matrix. Example: $\begin{bmatrix} 1& 2&3 \ 4&5 &6 \ 7&8 &9 \end{bmatrix}^{T} = \begin{bmatrix} 1&4 &7 \ 2&5 &8 \ 3&6 &9 \end{bmatrix}$ ie. column 1 become row 1, column 2 becomes row 2, etc.

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Question

$C_{ \mathbb{R}}$ refers to the set of all functions with domain $\mathbb{R}$ and range a subset of $\mathbb{R}$ .

Define the transformation $T: C_{ \mathbb{R}} \rightarrow \mathbb{R}$ to be

$T(f) = \sum_{j=1}^{100} f'(j)$

True or false: is a linear transformation.

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Answer

For to be a linear transformation, it must hold that

and

for all in the domain of and and for all scalar .

Let $f, g \in C_{ \mathbb{R}}$ .

$T(f) = \sum_{j=1}^{100} f'(j)$ and $T(g) = \sum_{j=1}^{100} g'(j)$ , so

$T(f)+ T(g)= \sum_{j=1}^{100} f'(j)+ \sum_{j=1}^{100} g'(j)$

By the sum rule for finite sequences,

$T(f)+ T(g)= \sum_{j=1}^{100}[ f'(j) + g'(j)]$

By the derivative sum rule,

$T(f)+ T(g)= \sum_{j=1}^{100}( f '+g')(j)$

$T(f)+ T(g)= \sum_{j=1}^{100}( f +g)'(j)$

The first condition is met.

Let $f \in C_{ \mathbb{R}}$ and be a scalar.

$T(f) = \sum_{j=1}^{100} f'(j)$

$c T(f) = c\sum_{j=1}^{100} f'(j)$

By the scalar product rule for finite sequences,

$c T(f) = \sum_{j=1}^{100}c f'(j)$

By the scalar product rule for derivatives,

$c T(f) = \sum_{j=1}^{100}(c f)'(j)$

The second condition is met.

is a linear transformation.

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Question

$C_{[0, 2\pi]}$ refers to the set of all functions that are continuous on $[0,2\pi]$ .

Define the linear mapping $T: C_{[0, 2\pi]}\rightarrow \mathbb{R}$ as follows:

$T(f)= \int_{0}^{2 \pi} f(x) \textup{ d}x$

True or false: $f(x) = \cos x + \sin x$ is in the kernel of .

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Answer

The kernel of a linear transformation is the subset of the domain of that maps into the zero of its range. It follows that $f \in \ker (T)$ if and only if

$T(f)= \int_{0}^{2 \pi} f(x) \textup{ d}x = 0$

To determine whether this is true or false, evaluate the integral:

$T(f)= \int_{0}^{2 \pi} (\cos x + \sin x) \textup{ d}x = 0$

$=\ \left.\begin{matrix} \sin x - \cos x \ ; \end{matrix}\right|\ \begin{matrix} 2 \pi \ 0 \end{matrix}$

$= (\sin 2\pi - \cos 2 \pi) - (\sin 0 - \cos 0)$

Therefore, $f \in \ker (T)$ .

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Question

$C_{[0, \pi]}$ refers to the set of all functions that are continuous on $[0, \pi]$ .

Define the linear mapping $T: C_{[0, \pi]}\rightarrow \mathbb{R}$ as follows:

$T(f)= \int_{0}^{ \pi} f(x) \textup{ d}x$

True or false: $f(x) = \cos x + \sin x$ is in the kernel of .

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Answer

The kernel of a linear transformation is the subset of the domain of that maps into the zero of its range, so, by definition, $f \in \ker (T)$ if and only if

$T(f)= \int_{0}^{ \pi} f(x) \textup{ d}x = 0$

To determine whether this is true or false, evaluate the integral:

$T(f)= \int_{0}^{ \pi} (\cos x + \sin x) \textup{ d}x = 0$

$=\ \left.\begin{matrix} \sin x - \cos x \ ; \end{matrix}\right|\ \begin{matrix} \pi \ 0 \end{matrix}$

$= (\sin \pi - \cos \pi) - (\sin 0 - \cos 0 )$

$T(f) \ne 0$ , so $f \notin \ker (T)$ .

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Question

True or false: If $T: V \rightarrow W$ is a linear mapping, and is a vector space, then is a subspace of .

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Answer

For example, if is the space of all vectors in $\mathbb{R}^2$ of the form , and is the space of all vectors in $\mathbb{R}^2$ the form , then $T: (a_1,a_2) \rightarrow (a_1,0)$ is a linear mapping, but is not a subset of , let alone a subspace of .

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Question

True or false: The identity mapping $I:V \rightarrow V$ , is also considered a linear mapping, regardless of the vector space .

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Answer

Verifying the conditions for a linear mapping, we have

$I[a\mathbf{x}+b]=a\mathbf{x}+b = aI[\mathbf{x}]+b.$

Hence the identity mapping is closed under vector addition and scalar multiplication, and is therefore a linear mapping.

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Question

$P_{3}$ is the set of all real polynomials with degree 3 or less.

Define the linear mapping $T: P_{3}\rightarrow \mathbb{R}$ as follows:

$T(P(x)) \rightarrow P(1)$

Is this linear mapping one-to-one and onto?

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Answer

$T: P_{3}\rightarrow \mathbb{R}$ is a linear mapping of a four-dimensional vector space into a one-dimensional vector space; it cannot be one-to-one.

is onto if, for every element in its codomain, which here is $\mathbb{R}$ , there exists at least one in the domain $P_{3}$ so that .

For each , we can choose the constant polynomial , so each has at least one domain element that maps into it.

is onto but not one-to-one.

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Question

Find the Eigen Values for Matrix .

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

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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

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

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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 $\mathbf{v} = \begin{bmatrix} -3\1 \end{bmatrix}$

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

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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 $\mathbf{v} = \begin{bmatrix} 1\ 1\end{bmatrix}$ ?

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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

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

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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 $\mathbf{v} = \begin{bmatrix} 2\ 3\end{bmatrix}$

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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

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

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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

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

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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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