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Calculating the Kernel, dimension of linear equations, real numbers and galois field

1

$begingroup$

Given these problems below, how would one calculate the result?

By intuition, I have managed to solve two of them but cannot crack the last one.

Btw, I am not sure that the approach of my intuition is the right one.

I am interested in learning about how to solve these (formulas and approach).

Problems Below:

1) Let $f: ℝ^7 rightarrow ℝ^4$ be a surjective (onto) linear function. What is the dimension of $Ker space f$? (answer = 3)

2) Let $f: GF(2)^8 rightarrow GF(2)^9$ be a linear funktion with $dim space Im space f=5$. How many vectors are in $Ker f$? (answer = 8)

3) Let $f: ℝ^7 rightarrow ℝ^14$ be a injective (one-to-one) linear function. Determine $dim space Im space f$? (answer = 7)

Could determine the answers for problem 1 and 3 by intuition (but not by formula). Here is the intuition below:

1 - Intuition: The function is surjective so the dimension must be $7 - 4 = 3$

3 - Intuition: The function is injective so the dimension must be $14 - 7 = 7$

(First question on site, so open for constructive feedback regarding the question.)

linear-algebra matrices matrix-equations matrix-rank

share|cite|improve this question

asked Apr 1 at 19:07

ms99ms99

204

$endgroup$

add a comment | 

1

$begingroup$

Given these problems below, how would one calculate the result?

By intuition, I have managed to solve two of them but cannot crack the last one.

Btw, I am not sure that the approach of my intuition is the right one.

I am interested in learning about how to solve these (formulas and approach).

Problems Below:

1) Let $f: ℝ^7 rightarrow ℝ^4$ be a surjective (onto) linear function. What is the dimension of $Ker space f$? (answer = 3)

2) Let $f: GF(2)^8 rightarrow GF(2)^9$ be a linear funktion with $dim space Im space f=5$. How many vectors are in $Ker f$? (answer = 8)

3) Let $f: ℝ^7 rightarrow ℝ^14$ be a injective (one-to-one) linear function. Determine $dim space Im space f$? (answer = 7)

Could determine the answers for problem 1 and 3 by intuition (but not by formula). Here is the intuition below:

1 - Intuition: The function is surjective so the dimension must be $7 - 4 = 3$

3 - Intuition: The function is injective so the dimension must be $14 - 7 = 7$

(First question on site, so open for constructive feedback regarding the question.)

linear-algebra matrices matrix-equations matrix-rank

share|cite|improve this question

asked Apr 1 at 19:07

ms99ms99

204

$endgroup$

add a comment | 

$begingroup$

Given these problems below, how would one calculate the result?

By intuition, I have managed to solve two of them but cannot crack the last one.

Btw, I am not sure that the approach of my intuition is the right one.

I am interested in learning about how to solve these (formulas and approach).

Problems Below:

1) Let $f: ℝ^7 rightarrow ℝ^4$ be a surjective (onto) linear function. What is the dimension of $Ker space f$? (answer = 3)

2) Let $f: GF(2)^8 rightarrow GF(2)^9$ be a linear funktion with $dim space Im space f=5$. How many vectors are in $Ker f$? (answer = 8)

3) Let $f: ℝ^7 rightarrow ℝ^14$ be a injective (one-to-one) linear function. Determine $dim space Im space f$? (answer = 7)

Could determine the answers for problem 1 and 3 by intuition (but not by formula). Here is the intuition below:

1 - Intuition: The function is surjective so the dimension must be $7 - 4 = 3$

3 - Intuition: The function is injective so the dimension must be $14 - 7 = 7$

(First question on site, so open for constructive feedback regarding the question.)

linear-algebra matrices matrix-equations matrix-rank

share|cite|improve this question

asked Apr 1 at 19:07

ms99ms99

204

$endgroup$

share|cite|improve this question

asked Apr 1 at 19:07

ms99ms99

204

share|cite|improve this question

asked Apr 1 at 19:07

ms99ms99

204

asked Apr 1 at 19:07

ms99ms99

204

asked Apr 1 at 19:07

ms99ms99

204

1 Answer
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1

$begingroup$

All you need here is the dimension theorem: if $f:Uto V$ is any linear map, then
$$dimker f+dimmathrmim f,=, dim U$$

Choose bases $u_1,dots, u_k$ for $ker f$ and $v_1,dots, v_r$ for $mathrmim f$ and arbitrary preimages $w_j$ of $v_j$. Then show $u_1,dots, u_k, w_1,dots, w_r$ is a basis of $U$.

This resolves your uncertainty for a) and c), and gives $dimker f=3$ for b).

So that $ker fcong GF(2)^3$ (by coordinating the vectors) which has $2^3$ elements.

share|cite|improve this answer

answered Apr 1 at 22:49

BerciBerci

62k23776

$endgroup$

add a comment | 

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1

$begingroup$

All you need here is the dimension theorem: if $f:Uto V$ is any linear map, then
$$dimker f+dimmathrmim f,=, dim U$$

Choose bases $u_1,dots, u_k$ for $ker f$ and $v_1,dots, v_r$ for $mathrmim f$ and arbitrary preimages $w_j$ of $v_j$. Then show $u_1,dots, u_k, w_1,dots, w_r$ is a basis of $U$.

This resolves your uncertainty for a) and c), and gives $dimker f=3$ for b).

So that $ker fcong GF(2)^3$ (by coordinating the vectors) which has $2^3$ elements.

share|cite|improve this answer

answered Apr 1 at 22:49

BerciBerci

62k23776

$endgroup$

add a comment | 

1

$begingroup$

All you need here is the dimension theorem: if $f:Uto V$ is any linear map, then
$$dimker f+dimmathrmim f,=, dim U$$

Choose bases $u_1,dots, u_k$ for $ker f$ and $v_1,dots, v_r$ for $mathrmim f$ and arbitrary preimages $w_j$ of $v_j$. Then show $u_1,dots, u_k, w_1,dots, w_r$ is a basis of $U$.

This resolves your uncertainty for a) and c), and gives $dimker f=3$ for b).

So that $ker fcong GF(2)^3$ (by coordinating the vectors) which has $2^3$ elements.

share|cite|improve this answer

answered Apr 1 at 22:49

BerciBerci

62k23776

$endgroup$

add a comment | 

$begingroup$

All you need here is the dimension theorem: if $f:Uto V$ is any linear map, then
$$dimker f+dimmathrmim f,=, dim U$$

Choose bases $u_1,dots, u_k$ for $ker f$ and $v_1,dots, v_r$ for $mathrmim f$ and arbitrary preimages $w_j$ of $v_j$. Then show $u_1,dots, u_k, w_1,dots, w_r$ is a basis of $U$.

This resolves your uncertainty for a) and c), and gives $dimker f=3$ for b).

So that $ker fcong GF(2)^3$ (by coordinating the vectors) which has $2^3$ elements.

share|cite|improve this answer

answered Apr 1 at 22:49

BerciBerci

62k23776

$endgroup$

share|cite|improve this answer

answered Apr 1 at 22:49

BerciBerci

62k23776

answered Apr 1 at 22:49

BerciBerci

62k23776

answered Apr 1 at 22:49

BerciBerci

62k23776