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I am unsure of my answers here any confirmation on my answers, additional knowledge or input would be awesome.

a) All vectors of the form $(a,b,c)$ where $a+b+c = 0$ form a subspace in $R^3$ True or False? why?

I believe this is true since this would mean it contains the zero vector which is required for a subspace, but I am unsure is this right?

b) The set of all 3 by 3 matrices is a vector space of dimension 6? True or False? Why?

False, because this forms a system of equations in only 3 dimensions?

c) The set of all polynomials of degree 3 is a vector space. True or False?Why?

No clue on this one.

d) Any four vectors in $R^5$ are linearly dependent. True or False? Why?

True, because depending on what our 5th vector is it could change the whole problem to be linearly independent.

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    $\begingroup$ You need to work from the definitions. Your correct answers are poorly justified. You have a list of conditions that a set must respect to be a vector space, just go through that list in all examples. $\endgroup$
    – Git Gud
    Apr 29, 2014 at 21:49
  • $\begingroup$ This is $\ker [1,1,1]$. So you know it's a vector space, and it shouldn't be hard to determine other things about it. $\endgroup$
    – user142299
    Apr 29, 2014 at 21:52
  • $\begingroup$ Not sure what your saying here. $\endgroup$
    – Achilles
    Apr 29, 2014 at 21:56

3 Answers 3

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a) true

just use the axioms. suppose $(a,b,c)$ and $(d,e,f)$ satisfy your condition. $(a,b,c)+x*(d,e,f) = (a+xd,b+xe,c+xf)$ => $a+xd+b+xe+c+xf = (a+b+c)+x*(d+e+f) = 0$

b) false

it's a 9-dimensional vector space

c) false

because: $(x^3 + 1)$ is a degree 3 polynomial. $(-x^3 + x)$ is a degree 3 polynomial. but $(x^3+x) + (-x^3 + x) = (x+1)$ is not a degree 3 polynomial but a degree 1 polynomial.

d) false

the five unit vectors are independent. so, the first 4 of them are independent too. so, not every 4 vectors are dependent.

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    $\begingroup$ Regarding c): It is actually true. The polynomials do make up a vector space as $P_{3}$ includes the polynomials of the lower stages. If this were not true it would suffice to point out that the space has no zero element, as the degree of the zero polynomial is negative infinity. If you go through the definition of vector space keeping this in mind, you will find that all the requirements hold. $\endgroup$
    – user102184
    Apr 29, 2014 at 22:07
  • $\begingroup$ the set of all polynomials with degree <= 3 is a vector space. but the set of polynomials with degree = 3 doesn't include 0, and it doesn't include polynomials of degree 0, 1 or 2. since (x+1) has degree=1 it cannot have degree=3 at the same time. So it's not a vector-space. $\endgroup$
    – Michael
    Apr 29, 2014 at 22:09
  • $\begingroup$ Yes, I see know that you are indeed correct. I just saw degree 3 and assumed that OP meant the space $P_{3}$, which is standard. $\endgroup$
    – user102184
    Apr 29, 2014 at 22:13
  • $\begingroup$ The term "3-dim polynomial" makes no sense. Rather, you should say a "degree 3 polynomial". $\endgroup$ Apr 29, 2014 at 22:15
  • $\begingroup$ @BillDubuque: thanks. fixed it. $\endgroup$
    – Michael
    Apr 29, 2014 at 22:18
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The notion 'vector space' is a formalism where we can write things like $a\cdot {\bf u}+b\cdot{\bf v}$ where $a$ and $b$ are (real) numbers, and ${\bf u},\ {\bf v}$ can be anything: either geometrically vectors indeed, or polynomials, or equations, whatever that you can multiply by a number and add.

All that, the structure has to satisfy the familiar algebraic rules of these operations: see e.g. on wikipedia.


a) To check the zero vector is a good beginning. You also have to check that addition and (multiplication by real number) doesn't lead out from the given set (which is always true for linear expressions).

b) We have to use $9$ variables to write a generic $3\times 3$ matrix: $\pmatrix{a&b&c\\d&e&f\\g&h&i\\}$. The elements of the standard basis for the matrices are matrices which has exactly one $1$ entry and the rest are $0$.

c) Can you multiply a polynomial by a number? Can you add polynomials? Will the result stay in the given set (of degree exactly $3$ or less than or equal to $3$?)

d) $\Bbb R^5$ has dimension $5$, it means that there can be at most $5$ independent vectors. (There meet $5$ half-lines in the corner of a room in a $5$ dimensional space, each orthogonal to each other [hence linearly independent].)

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Stick to the definitions, in order for a subset $W$ of a vector space $V$ to be a subspace it must satisfy two conditions:

$(1)$ $\forall w \in W, cw \in W$ (Closure under scalar multiplication).

$(2)$ $\forall v,w \in W, v + w \in W$ (Closure under vector addition).

Condition $(1)$ tells us that $0 \in W$ if $W$ is a subspace because we can simply take $c = 0$, so if the zero vector is not in $W$ then you know $W$ can't be a subspace because it's not closed under scalar multiplication. However, the presecene of the zero vector doesn't guarantee that $W$ is a subspace.

I'll outline the sketch to the answer to (a) since your other questions aren't directly related to subspaces.

(a) Graphically this represents a plane through the origin (fun fact: any plane through the origin is automatically a subspace of $R^3$), you can show this by just following the definitions. That is to say consider a point on the plane show that any scalar multiple of that point will also be on the plane and then consider two points on the plane show that their sum must also be on the plane.

To test understanding try answering the following question: True or False: All vectors of the form $(a,b,c)$ where $abc = 0$ forms a subspace of $R^3$.

False $(0,0,1) \in W$ and $(1,1,0) \in W$, but their sum $(1,1,1) \notin W$, since $W$ is not closed under vector addition it can't be a subspace.

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  • $\begingroup$ Thank you, im giving it a try now. $\endgroup$
    – Achilles
    Apr 29, 2014 at 22:32

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