# T/F Linear Alg ?s

1. May 10, 2009

### tas3113

I have a few T/F questions I would like to double check..

1. If A is a 3 x 4 matrix, then the kernel of A is a subspace of R^3.
I say this is false because since the kernel is a subspace of R^4, since there are 4 columns.

2. If A is a symmetric matrix, all of its eigenvalues are real and have multiplicity one.

3. If A is a Hermitian matrix, all of its eigenvectors are real.

4. If A is a square matrix with determinant zero, then its columns are linearly dependent.
I say this is true. Not sure why though. I'm thinking that there has to be a relationship between the columns since they will need to cancel out.

5. If A is a 4 x 3 matrix of rank 3, then its columns span R^3.
I say this is true because rank equals the number of pivots. Since there are 3 columns and 3 pivots, the basis is made of the 3 columns. Thus it spans R^3.

2. May 10, 2009

### jbunniii

If A is a 3 x 4 matrix, then A maps from R^m to R^n. What are m and n? Is the kernel a subspace of the "from" vector space or the "to" vector space? (Your answer is right, but I'm not sure about your justification.)

Try some examples. A diagonal matrix is a nice, simple example of a symmetric matrix. What can you say about the eigenvalues of a diagonal matrix?

Start by answering this question: What can you say about the eigenVALUES of a Hermitian matrix?

Can you list what facts you know must be true if you have a square matrix with determinant equal to zero?

If A is a 4 x 3 matrix, then A maps from R^m to R^n. What are m and n? Is the "span of the column space" a subspace of the "from" vector space or the "to" vector space? Then answer: what does "rank" refer to terms of the previous question? And also: if you have a subspace of R^n, does it follow that the subspace spans R^k for some k <= n?

3. May 10, 2009

### tas3113

2. the eigenvalues of a diagonal matrix must have multiplicity one. so therefore a symmetric matrix must have eigenvalues of multiplicity one. as for the real part, i want to say yes but don't know why.

True?

3. True?
all eigenvalues of a Hermitian matrix are real

4. Well for a 2 x 2, ad-bc = 0. well I guess you can have a matrix that's
(0, 0)
(1, 3)
and those columns are linearly independent. so false?

5. i'm still confused. i never really understood how this mapping, spanning and subspace thing works.

4. May 10, 2009

### jbunniii

Consider the 2x2 identity matrix

$$\left[ \begin{array}{ccc} 1 & 0 \\ 0 & 1 \end{array} \right]$$

What are its eigenvalues, and what are the multiplicities of those eigenvalues?

Also, part of the question is whether the eigenvalues of a symmetric matrix have to be REAL. Consider the 2x2 diagonal matrix

$$\left[ \begin{array}{ccc} i & 0 \\ 0 & i \end{array} \right]$$

What are the eigenvalues of this matrix?

Correct: the eigenvalues are real. Now, if you have a Hermitian matrix $$M$$, a real eigenvalue $$\lambda$$, and an eigenvector $$x \neq 0$$, satisfying

$$Mx = \lambda x$$

then does $$x$$ have to be real? (Hint: what if $$M$$ is not real?)

No, those columns are linearly dependent: the second column is three times the first column.

Here's a hint: the determinant of a matrix is zero if and only if the matrix is not invertible. Now, what can you say about the columns of a matrix that is not invertible?

OK, let's take it one step at a time.

A 4x3 matrix takes input vectors from R^3 (i.e., three-dimensional vectors) and produces output vectors in R^4 (i.e., four-dimensional vectors).

Let's not even worry about the rank of the matrix for now.

Each column of the matrix is a vector in R^4. Are there any circumstances under which a set of four-dimensional vectors can span R^3?

If I form a linear combination of vectors in R^4, the result is still in R^4. (It is a vector containing four elements.) Can I span R^3 with vectors of this size?

5. May 10, 2009

### tas3113

2. it has eigenvalue 1 with multiplicity 2 and eigenvectors (1,0) , (0,1) -- so it doesn't have to have multiplicity one
and i guess the values along the diagonal can be any value. real or imaginary. so this too is false.
so false

3. no, its eigenvectors can be imaginary. the Hermitian matrix will have/can have imaginary components. and if the eigenvalue is real, then x must be imaginary to have them equal each other. thus false

4. duh - i knew that. blanked out for a sec i guess.

so the determinant of a matrix is zero if and only if the matrix is not invertible.
well a property for an invertible matrix is - The columns of A are linearly independent. if it is not invertible, then the columns of A are linearly dependent? so true.

5. Each column of the matrix is a vector in R^4. Are there any circumstances under which a set of four-dimensional vectors can span R^3? Yes, linearly independent and have rank 3.
(1),(0),(0)
(0),(1),(0)
(0),(0),(1)
(0),(0),(0)
that spans R^3

If I form a linear combination of vectors in R^4, the result is still in R^4. (It is a vector containing four elements.) Can I span R^3 with vectors of this size?
yes?

6. May 10, 2009

### jbunniii

OK, great! All of the above is correct.

No, a vector in R^3 looks like

$$\left[ \begin{array}{c} a & b & c \end{array} \right]$$

whereas any linear combination of columns of a 4x3 matrix is a vector in R^4, which looks like

$$\left[ \begin{array}{c} w & x & y & z \end{array} \right]$$

so there's no way that vectors of this size can span R^3: they don't have the right number of elements!

What CAN happen is that the columns of a 4x3 matrix can span a SUBSPACE of R^4, such that the dimension of the subspace is 3 (this occurs if the rank of the matrix is 3). But that is NOT the same as spanning R^3! A subspace of R^4 that has dimension 3 might look something like all vectors of the form

$$\left[ \begin{array}{c} w & x & y & 0 \end{array} \right]$$

or all vectors of the form

$$\left[ \begin{array}{c} 0 & x & y & z \end{array} \right]$$

or so forth. Such a subspace is ISOMORPHIC to R^3, but it is not R^3.

7. May 10, 2009

### tas3113

ah. ok - so since there are only 3 columns and a rank of 3 then there is one row which does not have a variable? so then it does span R^3

but im also thinking..
it can be
$$\left|\begin{array}{ccc}w & w & w \\ x & x & x \\ 0 & y & y \\ 0 & 0 & z\end{array}\right|$$
that has a rank 3 but is spanning R^4 right? so then it would be false?

8. May 11, 2009

### jbunniii

For any 4x3 matrix M, the fact that there are only three columns means that the rank is at most 3. Why? Because the rank is simply the number of linearly independent columns in the matrix. (Or, equivalently, the rank is the dimension of the subspace spanned by the columns of M, which is simply range(M).)

Now for your particular 4x3 matrix M, you are given that rank(M) is 3, the maximum possible.

This means that it has three linearly independent columns. How big is each column vector? It is a 4x1 vector, i.e., it is an element of R^4. If I form any linear combination of the columns, I get an element of R^4. I do NOT get an element of R^3 (a 3x1 vector). Therefore there is no way, no how, that the columns can span R^3. The columns aren't even elements of R^3! Thus the answer has to be "false." Does that make sense?

Now, what exactly DO the column vectors of M span? Well, by definition they span range(M), which is a subspace of R^4. Why is it a subspace of R^4? Because it's simply the set of all linear combinations of the columns, and the columns live in R^4. Therefore range(M) is a subspace of R^4. What is the dimension of range(M)? It is precisely the rank of M, which is 3. Therefore, the columns of M span this: "a three-dimensional subspace of R^4." Note that this is absolutely NOT the same as spanning R^3.

What IS a three-dimensional subspace of R^4?

Well, it's a collection of 4-dimensional vectors that have only three degrees of freedom, for example the set of all 4-dimensional vectors of the following form:

$$\left[ \begin{array}{c} a & b & c & 0\end{array} \right]$$

So great, is a 3-dimensional subspace of R^4 simply the set of all 4-dimensional vectors that have a "0" in a given slot? Not necessarily: another example is the set of all 4-dimensional vectors of the following form:

$$\left[ \begin{array}{c} a & b & c & c\end{array} \right]$$

Please work very carefully through everything above and make sure you understand it all, because this stuff is really fundamental to linear algebra. If you have any questions at all, please feel free to reply with them!

Last edited: May 11, 2009
9. May 11, 2009

### tas3113

ok. im going to read this through a few more times to understand it. ill make sure to ask if i have any questions. thanks for your help on this problem