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Decide whether the map is linear

"Consider

\psi : R^{3} \mapsto R^{2}

\psi (a , b, c) =

\begin{pmatrix} 7a + 6b - 3c \\ -9a + 8b - 3c \end{pmatrix}

Decide whether

\psi

is a linear map."

How do you work this out?

Reply 1

nuodai

There's not much advice I can really give other than... check whether it satisfies the conditions for being a linear map.

Reply 2

claret_n_blue

Original post by nuodai

There's not much advice I can really give other than... check whether it satisfies the conditions for being a linear map.

The two conditions (from wikipedia) are:

f (\vec{x} + \vec{y}) = f(\vec{x}) + f(\vec{y})

and

f(\alpha x) = \alpha f(x)

I don't get how to use them.

I don't get what they mean by f(x).

Reply 3

nuodai

Original post by claret_n_blue

The two conditions (from wikipedia) are:

f (\vec{x} + \vec{y}) = f(\vec{x}) + f(\vec{y})

and

f(\alpha x) = \alpha f(x)

I don't get how to use them.

I don't get what they mean by f(x).

Here

\vec{x}

is interpreted to be a vector

(a,b,c)

. So for example

\alpha(a,b,c) = (\alpha a, \alpha b, \alpha c)

.

So you need to check whether or not

f(a, b, c) + f(p, q, r) = f(a + p, b + q, c + r)

and

\lambda f(a,b,c) = f(\lambda a, \lambda b, \lambda c)

both hold. If they do then it's linear; if they don't then it isn't.

Reply 4

claret_n_blue

Original post by nuodai

Here

\vec{x}

is interpreted to be a vector

(a,b,c)

. So for example

\alpha(a,b,c) = (\alpha a, \alpha b, \alpha c)

.

So you need to check whether or not

f(a, b, c) + f(p, q, r) = f(a + p, b + q, c + r)

and

\lambda f(a,b,c) = f(\lambda a, \lambda b, \lambda c)

both hold. If they do then it's linear; if they don't then it isn't.

I don't get what you mean by

f(a, b, c) + f(p, q, r) = f(a + p, b + q, c + r)

How would I check that? Surely that hold for everything, regardless of whatever numbers there are in it?

With my question, what I did was:

\begin{pmatrix} 7a + 6b - 3c \\ -9a + 8b - 3c \end{pmatrix}

So therefore

7a + 6b - 3c - 9a + 8b - 3c = -2a + 14b - 6c

That shows us what

f(a, b, c) + f(p, q, r)

equals.

Then:

f(a + p, b + q, c + r) = (7 - 9), (6 + 8), (-3 - 3) = (-2, 14, -6)

which is the same for the first one, so the addition axiom holds.

For the multiplication one, do I simply say that:

\lambda \begin{pmatrix} 7 & -9 \\6 & 8 \\-3 & -3 \end{pmatrix} =

\begin{pmatrix} \lambda7 & \lambda(-9)\\ \lambda6 & \lambda8\\ \lambda(-3) & \lambda(-3) \end{pmatrix}

so that axiom holds and therfore

\psi

is a linear map.

Sorry if they're dumb questions, I'm just struggling a bit getting my head around this. I'm sure once I've done one I should be able to understand it a bit more.

(edited 12 years ago)

Reply 5

nuodai

What on earth...? Something is fundamentally wrong with your understanding of this question. What's this related to? Do you have lecture notes on this, etc?

Reply 6

claret_n_blue

Original post by nuodai

What on earth...? Something is fundamentally wrong with your understanding of this question. What's this related to? Do you have lecture notes on this, etc?

I have notes but they say exactly the same thing as Wikipedia and so I'm struggling to understand.

Just to let you know. This question is a practice question and I already know the answer, this is a linear map. I'm just struggling as to how they got to that conclusion so I can actually do it for the real question.

This is an example we did:

"

\psi R^{2} \mapsto R^{2} , \psi (x, y) = (2x, 2y)

1st axiom:

\psi ( (x_1, y_1) + (x_2, y_2) ) = \psi ( (x_1 + x_2), (y_1 + y_2) ) = ( 2(x_1 + x_2), 2(y_1 + y_2) )

\psi (x_1, y_1) + \psi (x_2, y_2) = (2x_1, 2y_1) + (2x_2, 2y_2) = = (2x_1 + 2x_2, 2y_1 + 2y_2) = ( 2(x_1 + x_2), 2 (y_1 + y_2) )

So the first axiom holds.

2nd axiom:

\psi ( \lambda (x,y)) = \psi (\lambda x, \lambda y) = (2\lambda x, 2\lambda y)

\lambda \psi (x, y) = \lambda (2x, 2y) = (2\lambda x, 2\lambda y)

So the second axiom holds.

I understand all this because they told me exactly what psi was and what it equals so I can follow the steps. But in my question, they've given it to me in a matrix and instead of going from R² to R², it goes from R³ to R². That throws me a little.