Consider this matrix and comment on its surjectivity and injectivity

[flyingpig]

Homework Statement

$$\begin{bmatrix} 1 & 0 & 0 & 6\\ 0& 1 & 0 & 2\\ 0& 0 & 1& 3\\ 0 & 0& 0& 0 \end{bmatrix}$$

Consider the matrix (above) and comment whether it is "one to one" and "onto"







Book solution

My book says: Because not every column is a pivot column, therefore there exists a nontrivial solution which therefore is not one to one. Also, since the matrix does not have a pivot column in every row, the column matrix does not span R⁴ and therefore it is not onto

The Attempt at a Solution

I don't understand the book. So what if it doesn't have a pivot column in every row? This is a coefficient matrix, what if the augmented column has a 0 at the "corner" of the matrix? Wouldn't that imply there must only be one solution and hence the trivial solution?

Since there exists (or at least could be) one solution, then it must be one to one. Why did the book ignore this possibility?

I am also guessing since this matrix does not Span R⁴, it cannot be onto

 

[Mark44]

Homework Statement

$$\begin{bmatrix} 1 & 0 & 0 & 6\\ 0& 1 & 0 & 2\\ 0& 0 & 1& 3\\ 0 & 0& 0& 0 \end{bmatrix}$$

Consider the matrix (above) and comment whether it is "one to one" and "onto"







Book solution

My book says: Because not every column is a pivot column, therefore there exists a nontrivial solution which therefore is not one to one. Also, since the matrix does not have a pivot column in every row, the column matrix does not span R⁴ and therefore it is not onto

The Attempt at a Solution

I don't understand the book. So what if it doesn't have a pivot column in every row? This is a coefficient matrix, what if the augmented column has a 0 at the "corner" of the matrix?

That's not what you're calling an augmented column, which would be the column of constants on the right sides of the equations this matrix would represent.

This matrix represents a system of three equations in four unknowns. Since the matrix is in RREF form, I can see that there is one free variable, and that the other three variables are defined in terms of this free variable.

Based on the book's answer, this is the correct interpretation of what this matrix means (that it is not an augmented matrix).

If we call the variables x, y, z, and w, the RREF form of the matrix says that x = -6w, y = -2w, z = -3w, and w = itself. Since w is arbitrary, there are an infinite number of solutions to the equation Ax = 0, where A is your 4x4 matrix.

Wouldn't that imply there must only be one solution and hence the trivial solution?

Since there exists (or at least could be) one solution, then it must be one to one. Why did the book ignore this possibility?

I am also guessing since this matrix does not Span R⁴, it cannot be onto

Saying that a 4x4 matrix does or does not span R⁴ makes no sense at all. It does make sense to say whether the columns or rows span R⁴.

 

[flyingpig]

What does it even mean for a row to span some Rⁿ, I actually thought they meant column because I didn't think row spanning exists.

 

[Mark44]

Suppose you had this matrix:
$$A = \begin{bmatrix}1 & 0 & 0 & 0\\ 0& 1 & 0 & 0 \\ 0& 0 & 1& 0\end{bmatrix}$$

This matrix could be viewed as the representation of a transformation from R⁴ to R³.

The columns, viewed as vectors in R³, span R³. Since there are four columns, there are too many vectors for them to be linearly independent, so the four vectors don't form a basis for R³.

The rows, viewed as vectors in R⁴, can't span R⁴, because there are only three of them. These three vectors are linearly independent, so they span a three-dimensional subspace of R⁴. This subspace is the image or range of the transformation. Every vector v such that v = Au is a linear combination of the rows of A.

Another subspace worth mentioning is the nullspace or kernel of A. Every vector u such that Au = 0 is in the kernel of A. Solving the equation Au = 0 gives x = 0, y = 0, z = 0, and w is arbitrary. This means that every vector in ker(A) is a multiple of
$$\begin{bmatrix} 0\\ 0 \\ 0 \\ 1\end{bmatrix}$$

 

[flyingpig]

Suppose you had this matrix:
$$A = \begin{bmatrix}1 & 0 & 0 & 0\\ 0& 1 & 0 & 0 \\ 0& 0 & 1& 0\end{bmatrix}$$

This matrix could be viewed as the representation of a transformation from R⁴ to R³.

What do you mean from R⁴ to R³? Surely it cannot be because there are four vectors right...?

The columns, viewed as vectors in R³, span R³. Since there are four columns, there are too many vectors for them to be linearly independent, so the four vectors don't form a basis for R³.

Hold on, does that mean in general that if a set is linearly independent, they can span a "new" dimension?

Let me try to elaborate

Let's say I have an n x m matrix and of course m < n. This set does not have a zero vector, none of them are multiple scalars of each other, and none of them can be written as a linear combination of each other (I am trying to rule out every possibities of being linearly dependent here, so if i miss one pretend I added it already) and hence this set is a linearly independent set.

So if this matrix has say n entries and rows, then does that mean regardless of what m is, this set can span Rⁿ

The rows, viewed as vectors in R⁴, can't span R⁴, because there are only three of them.

How would you even interpret them...?

Ex. $$\begin{bmatrix} 1 & 2 &3 & 4\\ 2& 3 &4 & 5 \end{bmatrix}$$

I read them as x + y + 3z = 4 and 2x + 3y + 4z = 5

But when you say row viewed as vectors do you mean

x + 2y + 3z = 4 and 2x + 3y + 4z = 5?

These three vectors are linearly independent, so they span a three-dimensional subspace of R⁴. This subspace is the image or range of the transformation. Every vector v such that v = Au is a linear combination of the rows of A.

Another subspace worth mentioning is the nullspace or kernel of A. Every vector u such that Au = 0 is in the kernel of A. Solving the equation Au = 0 gives x = 0, y = 0, z = 0, and w is arbitrary. This means that every vector in ker(A) is a multiple of
$$\begin{bmatrix} 0\\ 0 \\ 0 \\ 1\end{bmatrix}$$

I am sorry, but beyond this point, I have no idea what you are talking about LOL. I am using David Lays LA book if that helps...

 

[Mark44]

What do you mean from R⁴ to R³? Surely it cannot be because there are four vectors right...?

The matrix I gave represents a map from R⁴ to R³. Take any vector x in R⁴. Form the product Ax. The result is a vector in from R³.

Hold on, does that mean in general that if a set is linearly independent, they can span a "new" dimension?

If a set of m vectors in, say, R⁴, is linearly independent, these vectors span a subspace of dimension m, where 1 <= m <= n. I am assuming that your set has at least one nonzero vector in it.

Let me try to elaborate

Let's say I have an n x m matrix and of course m < n. This set

What set? You have a matrix. Are you talking about the columns, as vectors in R^m, or the rows, as vectors in Rⁿ.

Please be more specific and try to minimize the use of vague pronouns such as "it", "they", and so on.

does not have a zero vector, none of them are multiple scalars of each other, and none of them can be written as a linear combination of each other (I am trying to rule out every possibities of being linearly dependent here, so if i miss one pretend I added it already) and hence this set is a linearly independent set.

So if this matrix has say n entries and rows, then does that mean regardless of what m is, this set can span Rⁿ

The only way a matrix can have n entries and n rows is for it to be a column vector.

How would you even interpret them...?

Ex. $$\begin{bmatrix} 1 & 2 &3 & 4\\ 2& 3 &4 & 5 \end{bmatrix}$$

I read them as x + y + 3z = 4 and 2x + 3y + 4z = 5

You are assuming that this matrix is an augmented matrix, and there is no justification for this assumption.

This matrix could just as easily be the matrix for this system of equations:
x + 2y + 3z + 4w = 0
2x + 3y + 4z + 5w = 0

But when you say row viewed as vectors do you mean

x + 2y + 3z = 4 and 2x + 3y + 4z = 5?

No, not at all.
The rows in your matrix are vectors in R⁴; namely
<1, 2, 3, 4> and <2, 3, 4, 5>

The columns are vectors in R²; namely
$$\begin{bmatrix}1\\2\end{bmatrix}, \begin{bmatrix}2\\3\end{bmatrix},\begin{bmatrix}3\\4\end{bmatrix}, \text{and} \begin{bmatrix}4\\5\end{bmatrix}$$

I am sorry, but beyond this point, I have no idea what you are talking about LOL. I am using David Lays LA book if that helps...

I'm not convinced you understood what I was saying before that point, either.

 

[flyingpig]

No, not at all.
The rows in your matrix are vectors in R⁴; namely
<1, 2, 3, 4> and <2, 3, 4, 5>

Sorry for the late response, but the lag was insane yesterday and the day before...

But isn't <1,2,3,4> implying we are in R⁴? I thought the column vectors determine what R we are in.

 

[Mark44]

Sorry for the late response, but the lag was insane yesterday and the day before...

Yeah, it's been pretty bad since Saturday, but seems to be back to normal now.

But isn't <1,2,3,4> implying we are in R⁴? I thought the column vectors determine what R we are in.

Your question doesn't make much sense. We aren't "in" anyone place. This matrix -
$$A = \begin{bmatrix}1 & 2 &3 & 4\\ 2& 3 &4 & 5\end{bmatrix}$$

is a linear transformation from R⁴ to R². IOW, this matrix acts like a function that takes a vector in R⁴ and maps it to a vector in the plane (R²).

The domain of this transformation is all of R⁴: you can multiply any vector in R⁴ by this matrix. The rows, being that there are only two of them (and that they are linearly independent), span a two-dimensional subspace of R⁴. That doesn't mean that they are vectors in R². On the contrary, they have four components.

I haven't worked it out, but I suspect that the range of this transformation is all of R², meaning that no matter what vector v = <a, b> you pick, there is some vector u in R⁴ such that Au = v.

 

[flyingpig]

No I was referring to the "row vector" thing

 

[Mark44]

What's your question? I thought I had covered what you asked about.