Algebra II 189-251B Instructor: Dr. E. Goren.

Assignment 5

To be submitted by March 2, 12:00

Note: It is recommended that you use Maple or some other software to check, or even find, the results. However, the solution must always be a complete proof, independent of computer calculations.

1. Let

W_{1} = Span ({(1, 1, 0, 0), (0, 1, 1, −1)})
W_{2} = Span ({(3, 2, 1, 0), (4, 4, 2, −1)})

Find a system of homogeneous linear equations such that W_{1} is their solutions. The same for W_{2}. Find then W_{1} ∩ W_{2}.

2. (i) Compute the rank of the following system of linear equations, by finding the reduced echelon form.

5x_{1} + x_{2} + 3x_{3} + 2x_{4} + x_{5} = b_{1}
x_{1} + x_{3} + x_{4} = b_{2}
7x_{1} + 2x_{2} + 3x_{3} + x_{4} + 2x_{5} = b_{3}
−x_{1} − x_{2} + x_{3} + 2x_{4} − x_{5} = b_{4}

(ii) Prove that at least for one of the vectors

(\begin{matrix}b_{1} \\ b_{2} \\ b_{3} \\ b_{4} \end{matrix}) = \{(\begin{matrix}1 \\ 0 \\ 0 \\ 0 \end{matrix}) , (\begin{matrix}0 \\ 1 \\ 0 \\ 0 \end{matrix}) , (\begin{matrix}0 \\ 0 \\ 1 \\ 0 \end{matrix})\}

we can’t solve the system. (iii) Prove that there exists a non-trivial solution for the homogeneous system satisfying also x_{1} = x_{4}.

3.

(1) Prove that the determinant of an n×n matrix A = (a_{ij}) is non-zero if and only if the columns of A,

(a_{11} ; a_{21} ; ⋮ ; a_{n1}) , …, (a_{1n} ; a_{2n} ; ⋮ ; a_{nn}) ,

form a basis for 𝔽^{n}.

(2) Generalize this as follows. Let A = (a_{ij}) be an m × n matrix. Prove that rank(A) = k if and only if there exists a k × k sub-determinant of A that is not zero, and every (k + 1) × (k + 1) sub-determinant is zero.

By a k × k sub-determinant we mean the following. We choose k columns j_{1} < j_{2} < ⋯ < j_{k} among the n columns (so k is assumed ≤ n), and we choose k rows i_{1} < i_{2} < ⋯ < i_{k} among the m rows (so k is assumed ≤ m as well). We then look at the k × k matrix (a_{i_{ℓ}j_{m}}). Its determinant is what we call a k × k sub-determinant.

Example: Consider the matrix we tortured at the lectures

(\begin{matrix}3 & 2 & 0 \\ 1 & 1 & 1 \\ 2 & 1 & −1 \end{matrix})

We know it has rank 2. We note that the only 3×3 determinant is 0. It has nine 2×2 sub-determinants. Some of them are

det (\begin{matrix}3 & 2 \\ 1 & 1 \end{matrix}) , det (\begin{matrix}3 & 2 \\ 2 & 1 \end{matrix}) , det (\begin{matrix}2 & 0 \\ 1 & −1 \end{matrix}) , det (\begin{matrix}3 & 0 \\ 1 & 1 \end{matrix}) .

The determinants are 1, −1, −2, 3. Note that over every field the first determinant is non-zero. This shows that the matrix has rank 2. Note also that over the field ℤ/2ℤ the before last determinant is zero, and over ℤ/3ℤ the last determinant is zero. This shows that if some k × k sub-determinant is zero this need not hold for other k × k sub-determinants.

4. Give two more proofs of the following theorem Theorem 1. Let A be an m × n matrix. Then

rank_{c}(A) = rank_{r}(A).

(1) By applying Exercise 3 and properties of determinants.

(2) By completing the following argument. Write

A = ( R_{1} ; R_{2} ; ⋮ ; R_{m} ) .

If r = rank_{r}(A) then there exist vectors S_{1}, …, S_{r} in 𝔽^{n} such that for suitable k_{ij}

A = ( R_{1} ; R_{2} ; ⋮ ; R_{m} ) = ( ⋮  ⋮_{k_{m1} ... k_{mr}}^{k_{11} ... k_{1r}} ) ( S_{1} ; S_{2} ; ⋮ ; S_{r} ) .

(Explain why!). Now interpret each matrix as a linear transformation and see what you can say about the column rank!