(a) Find a homogeneous linear system of two equations in three unknowns whose solution space con

(a) Find a homogeneous linear system of two equations in...

Key Concepts

Homogeneous Linear System

A homogeneous linear system is one in which all of the constant terms are zero. This ensures that the solution set always contains the trivial solution, and, in fact, the entire solution set forms a subspace. Understanding these systems is fundamental as they reveal the structure of solutions in linear algebra.

Orthogonality and Orthogonal Complements

Orthogonality refers to the condition where two vectors have a dot product of zero, meaning they are perpendicular to each other. The orthogonal complement of a set of vectors consists of all vectors that are orthogonal to every vector in the set, and this concept is widely used in decomposing spaces and studying projections.

Null Space of a Matrix

The null space of a matrix is the set of all vectors that, when multiplied by the matrix, yield the zero vector. It represents the solution set of a homogeneous system and is a subspace whose dimension (the nullity) provides key insights into the degrees of freedom within the system.

Geometric Interpretation of Solution Spaces

Solution spaces of systems of linear equations can often be interpreted as geometric objects such as lines, planes, or hyperplanes. This geometric perspective is vital for visualizing the nature of the solution set, particularly in lower-dimensional spaces like R³.

General Solution and Parametric Form

Expressing a general solution in parametric form involves identifying free parameters that describe all possible solutions to the system. This method succinctly captures the structure of the solution set and directly reflects the dimension of the null space.

Rank-Nullity Theorem

The Rank-Nullity Theorem establishes a critical relationship in linear algebra: the number of variables in a system is equal to the sum of the rank of the matrix and the nullity (dimension of the null space). This theorem is a key tool in determining the existence and structure of solutions to a linear system.

Transcript

00:01 Hi, so for this exercise we have these two vectors, a and v, and we need to find a vector x on r3 that satisfy the condition that is orthogonal to these two vectors, and we need to express this vector as a system of linear equations, an homogeneous system of linear equations.

00:22 So basically we need that this vector should be orthogonal to a and to b, but the orthogality is equivalent to ask that x the inner product of this vector x with a is equal to zero and at the same time the inner product of the vector x with b is equal to zero so you can see that from this condition or tonality we can obtain two equations that will form a linear a system of linear an homogeneous system of linear equations okay so let's assume that our vector x so let x be equals to x1 x2 x3 so then this this is in their product corresponds to minus 3 x1 plus 2 x2 plus x3 equals to 0 and minus 2 x2 plus to x3 equals to zero.

01:37 So now we have this is this homogeneous system of linear equations and we can solve this final solution for this.

01:46 So what is the geometric interpretation of this? well basically each of these equations represent a plane that passed through the origin and the intersection of two planes will corresponds to a line.

02:08 If these two mechanisms, if these two, pass through the origin then this line will also pass through the origin so that's how you can interpret this system of linear equations but also you can interpret as a vector by the definition that is orthogonal to these two and v and that has a more deep geometric meaning that i'm going to discuss after finding the solution of this homogeneous system so let's first consider the the matrix to this system so we have minus 3 to minus 1 0 minus 2 to 0 0 0 then after just divided by 2 the second row so we take the second row on multiply by 1 half we obtained so we obtained the matrix minus 3 minus 2 minus 1 0 0 1 1 sorry here we multiply by minus 1 half so we obtain 1 minus 1 and 0 so from here we can obtain the solution and actually you can observe that we have 2 pi as expected because we have 3 unknowns just two equations so what happened is that x3 corresponds to a free variable so x3 will be defined as as t which is a real number any real number and then x2 will depend on t and x2 will be minus t and x1 will be equal to t so from this we can obtain a general solution that depends on the variable t and is given by t times minus 1 minus 1 and you can observe what happened here you can see that this x t is written as t times some vector and this is the structure of a line that pass through the origin so basically you have here r3 so one of the geometric interpretations of this is that you have here r3 and z and then you have here the origin and you have a vector so the vector is minus 1 minus 1 1 this is the vector v minus 1 minus 1 so more or less something like in this direction and go behind in this direction will be the line so this is the line that is represented by this parametric equation but as i mentioned before the geometric interpretation of this system and this solution is deeper than that and is because we have that this solution is obtained by linear by b b.

05:44 This solution is obtained by establishing that the solution will be orthogonal to these two vectors x and v so by establishing this we obtain a system of linear equation that we solve and then we obtain this solution here that corresponds to line that passed through the origin but basically that means that a and v span a plane okay so here you have the origin and then you have here a in this direction just for saying v these two vectors will span a plane in the space okay of course this plane is an infinite okay and what we have found here is the solution that is orthogonal to these two vectors but these two vectors span a plane so the line will be orthogonal to this plane this will corresponds to x t the solution and goes also below at infinity that intersects the plane through the origin but even more this vector this vector this this line is orthogonal to a and v and that means which are vectors at the span the plane so that means that this line is parallel to the normal of the plane so that is the full geometric interpretation of the system you have a line that pass through the origin where you have defined a plane that is defined by these two vectors a and v the span of those two vectors generated this plane and the solution that you have found is a line that passed through the origin as well and is orthogonal to the plane that implies that it is parallel to the normal so that's the full picture of this exercise so you can see that actually the algebra is not quite difficult but the geometric meaning is really full of things that is happening geometrically and that's the beautiful thing within linear algebra and geometry okay just to you the full picture of the geometric interpretation that what is happening here and then we need to solve the last part and the last part corresponds to verifying that we have defined a matrix that is minus 3 -2 -1 -0 minus 2 -0 and we need to verify that the solution that we have found x t equals to t times the vector minus 1, minus 1, 1, 1...

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