Question 6is:4 RGiven the map FR4R3 defined by f(x,y,z,w) = (0,0,0) Then rank ofA. 1B. 4O C.OD.3DE. 2

Given map is f:ℝ4→ℝ3 defined by f(x,y,z,w)=(0,0,0). we have to find the rank of f. The map f is…

Answered: Given the map FR4R3 defined by…

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Given map is $f : ℝ^{4} \to ℝ^{3}$ defined by $f (x, y, z, w) = (0, 0, 0)$ . we have to find the rank of f.

The map f is linear because:

I) let $(x, y, z, w) a n d (x_{1}, y_{1}, z_{1}, w_{1}) \in ℝ^{4}$ then

$f ((x, y, z, w) + (x_{1}, y_{1}, z_{1}, w_{1})) = f (x + x_{1}, y + y_{1}, z + z_{1}, w + w_{1}) = f (x, y, z, w) + f (x_{1}, y_{1}, z_{1}, w_{1}) = (0, 0, 0)$ and

II) $f (r (x, y, z, w)) = r f (x, y, z, w) = (0, 0, 0)$ , r is the element from the field.

To find the rank of f we use rank-nullity theorem of vector space.

Rank-nullity theorem:

Let $T : V \to W$ be a linear map and $V, W$ finite dimensional vector space then

$d i m (I m (T)) + d i m (k e r (T)) = d i m (V)$

Now , $d i m (I m (T)) = r a n k (T)$

therefore we can write, $r a n k (T) + d i m (k e r (T)) = d i m (V)$

According to given question we have , $T = f$ and $V = ℝ^{4}, W = ℝ^{3}$

Now apply rank-nullity theorem we get

$r a n k (f) + d i m (k e r (f)) = d i m (ℝ^{4})$

$r a n k (f) = 4 - d i m (k e r (f)), [d i m (ℝ^{4}) = 4]$ ............(1)

Now we find $d i m (k e r (f))$

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Given the map FR4R3 defined by f(x,y.z.w) = (0,0,0). Then rank of O A. 1- OB. 4 OC.O OD. 3 O E. 2